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NCERT Solutions for Class 12 Maths Chapter 7 - Integrals

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NCERT Solutions for Class 12 Maths Chapter 7 - Integrals| Free PDF Download

NCERT Solutions for Integrals Class 12 PDF can be downloaded now from Vedantu. The subject matter experts at Vedantu are deft in preparing tailor-made solutions for the Integrals Class 12 chapter taking into consideration all the needs of a student and providing help to manage their studies with efficiency. Their expertise in the education industry has made them highly capable to give unique and simple solutions to problems that may look daunting to students.


As you navigate the Integration Class 12 NCERT solutions on our platform, you'll discover that even the most challenging problems can be tackled with simplicity and ease. What sets our online solutions apart is our unwavering support—we don't leave you stranded in the midst of doubts. Instead, we provide continuous assistance, ensuring all your queries regarding Integrals NCERT solutions are addressed.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 7 - Integrals

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



But as you go through Integration Class 12 NCERT solutions on our portal, you would realize that even the most difficult problem can be handled with simplicity and ease. The bonus of accessing our online solutions is that we do not leave you in the middle of your doubts while going through our Integrals NCERT solutions but offer constant support in clearing all your queries. 



Integrals Chapter at a Glance - Class 12 NCERT Solutions

  • Integration is the inverse process of differentiation

Let $\frac{d}{d} F(x)=f(x)$. Then, we write $\int f(x) d x=F(x)+C$.

These integrals are called indefinite integrals or general integrals, $\mathrm{C}$ is called a constant of integration. All these integrals differ by a constant.

  • From the geometric point of view, an indefinite integral is a collection of family of curves, each of which is obtained by translating one of the curves parallel to itself upwards or downwards along the $y$-axis.

  • Some properties of indefinite integrals are as follows:

(i) $\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x$

(ii) For any real number $\dot{k} \cdot \int \dot{k} f(x) d x=\dot{k} \int f(x) d$ :

More generally, $\int\left[k_1 f_1(x)+k_2 f_2(x)+\ldots+k_n f_n(x)\right] d=k_1 \int f_1(x) d x+k_2 \int f_2(x) d x+\ldots+k_n \int f_n(x) d x$

  • Some standard integrals

(i) $\int x^n d: \frac{x^{n+1}}{n+1}+C, n \neq 1$.

Particularly, $\int d x=x+C$

(ii) $\int \cos x$ a $x=\sin x+C$

(iii) $\int \sin x \dot{x}=-\cos x+C$

(iv) $\int \sec ^2 x d x=\tan x+C$

(v) $\int \operatorname{cosec}^2 x \dot{d x}=-\cot x+C$

(vi) $\int \sec x \tan x d x-\sec x+C$

(vii) $\int \sec x \tan x a x=\sec x+C$

(viii) $\int \frac{d x}{\sqrt{1-x^2}}=\sin ^{-1} x+C$

(ix) $\int \frac{d x}{\sqrt{1-x^2}}=-\cos ^{-1} x+C$

(x) $\int \frac{d x}{1+x^2}=\tan ^{-1} x+C$

(xi) $\int \frac{d x}{1+x^2}=-\cot ^{-1} x+C$

(xii) $\int e^x d x-e^x+C$

(xiii) $\int a^x d x-\frac{a^x}{\log a}+C$

(xiv) $\int \frac{d}{x \sqrt{x^2-1}}=\sec ^{-1} x+C$

(xv) $\int \frac{d}{x \sqrt{x^2-1}}=-\operatorname{cosec}^{-1} x+C$

(xvi) $\int \frac{1}{x} d x=\log |x|+C$

  • Integration by partial fractions

Recall that a rational function is ratio of two polynomials of the form $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials in $x$ and $Q(x) \neq 0$.

If degree of the polynomial $P(x)$ is greater than the degree of the polynomial $Q(x)$, then we may divide $P(x)$ by $Q(x)$ so that $\frac{P(x)}{Q(x)}=T(x)+\frac{P(x)}{Q(x)}$, where $T(x)$ is a polynomial in $x$ and degree of $P_1(x)$ is less than the degree of $Q(x) . T(x)$ being polynomial can be easily integrated.

$\frac{P(x)}{Q(x)}$ can be integrated by expressing $\frac{P(x)}{Q(x)}$ as the sum of partial fractions of the following type:

\[\begin{aligned}& \frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}, \quad a \neq b \\& \frac{px+q}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2} \\& \frac{px^2+qx+r}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c} \\& \frac{px^2+qx+r}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-c} \\& \frac{px^2+qx+r}{(x-a)\left(x^2+bx+c\right)} = \frac{A}{x-a} + \frac{Bx+c}{x^2+bx+c}, \text{ where } x^2+bx+c \text{ cannot be factorized further.}\end{aligned}\]

  • Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. When the integrand involves some trigonometric functions, we use some well known identities to find the integrals. Using substitution technique, we obtain the following standard integrals.

(i) $\int \tan x \cos -\log |\sec x|+C$

(ii) $\int \cot x$ a $x-\log |\sin x|+C$

(iii) $\int \sec x d x-\log |\sec x+\tan x|+C$

(iv) $\int \cos x d x-\log |\operatorname{cosec} x+\cot x|+C$

  • Integrals of some special functions

(i) $\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$

(ii) $\int \frac{a}{a^2-x^2}-\frac{1}{2 a} \log \left|\frac{x+a}{x-a}\right|+C$

(iii) $\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C$

(iv) $\int \frac{a x}{\sqrt{x^2-a^2}}-\log \left|x+\sqrt{x^2-a^2}\right|+C$

(v) $\int \frac{\dot{a}}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$

(vi) $\int \frac{d}{\sqrt{x^2+a^2}}-\log \left|x+\sqrt{x^2+a^2}\right|+C$

  • Integration by parts

For given functions $f_1$ and $f_2$, we have $\int f_1(x) \cdot f_2(x) d x-f_1(x) \int f_2(x) d x-\int\left[\frac{d}{d x} f_1(x) \cdot \int f_2(x) d x\right] d$, i.e, the integral of the product of two functions = first function $x$ integral of the second function - integral of \{differential coefficient of the first function $x$ integral of the second function\}.

Care must be taken in choosing the first function and the second function

Obviously, we must take that function as the second function whose integral is well known to us.

One of the applications of the integration by parts is as follows:

/[\int e^x\left(f(x)+f'(x)\right) dx - \int e^x f(x) dx + C/]

  • Some special types of integral

(i) $\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C$

(ii) $\int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \operatorname{lng}\left|x+\sqrt{x^2-a^2}\right|+c$

(iii) $\int \sqrt{a^2-x^2} \dot{x}=\frac{x}{2} \sqrt{x^2-a^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C$

(iv) Integrals of the types $\int \frac{d x}{a x^2+b x+c}$ or $\int \frac{d x}{\sqrt{a x^2+b x+c}}$ can be transformed into standard form by expressing $a x^2+b x+c=a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]=a\left[\left(x+\frac{b}{2 a}\right)^2+\left(\frac{c}{a}-\frac{b^2}{4 a^2}\right)\right]$

(v) Integrals of the types $\int \frac{(p x+q) d x}{a^2+b x+c}$ or $\int \frac{(p x+q) d:}{\sqrt{a x^2+b x+c}}$ can be transformed into standard form by expressing $p x+q=A \frac{d}{d x}\left(a x^2+b x+c\right)+B=A(2 a x+b)+B$, where A and B are determined by comparing coefficients on both sides.

  • We have defined $\int_a^b f(x) a x$ as the area of the region bounded by the curve $y-f(x), a \leq x \leq b$, the $x$-axis and the ordinates $x=a$ and $x=b$. Let $x$ be a given point in $[a, b]$. Then $\int_a^b f(x) d x$ represents the Area function $A(x)$. This concept of area function leads to the Fundamental Theorems of Integral Calculus.

  • First fundamental theorem of integral calculus

Let the area function be defined by $A(x)-\int_a^x f(x)$ d for all $x \geq a$, where the function $f$ is assumed to be continuous on $[a, b]$. Then $A^{\prime}(x)=f(x)$ for all $x \in[a, b]$.

  • Second fundamental theorem of integral calculus

Let $f$ be a continuous function of $x$ defined on the closed interval $[a, b]$ and let $F$ be another function such that $\frac{d}{d} F(x)-f(x)$ for all $x$ in the domain of $f$, then $\int_a^b f(x) d x-[F(x)+C]_a^b-F(b)-F(a)$.

This is called the definite integral of $f$ over the range $[a, b]$, where $a$ and $b$ are called the limits of integration, $a$ being the lower limit and $b$ the upper limit.


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Exercises under NCERT Class 12 Maths Chapter 7 – Integrals

Chapter 7 of the Class 12 Maths NCERT textbook deals with Integrals. This chapter covers various topics such as indefinite integrals, definite integrals, methods of integration, integration using trigonometric identities, integration by substitution, integration by parts, integration using partial fractions, and miscellaneous integrals.


The exercises in this chapter are designed to help students gain a comprehensive understanding of integrals and their various applications.


Exercise 7.1 This exercise introduces the concept of indefinite integrals and teaches students how to find them.

Exercise 7.2 This exercise deals with the properties of indefinite integrals and helps students understand how to use these properties to solve problems.

Exercise 7.3 This exercise covers the concept of definite integrals and teaches students how to evaluate them using different methods such as the limit definition of definite integrals and the properties of definite integrals.

Exercise 7.4 This exercise teaches students how to evaluate definite integrals using substitution.

Exercise 7.5 This exercise deals with integration by parts and teaches students how to solve problems using this method.

Exercise 7.6 This exercise covers the concept of partial fractions and helps students understand how to use them to evaluate integrals.

Exercise 7.7 This exercise teaches students how to evaluate integrals using trigonometric identities.

Exercise 7.8 This exercise covers the concept of integration using trigonometric substitutions and helps students understand how to use this method to solve problems.

Exercise 7.9 This exercise teaches students how to evaluate integrals using the method of differentiation under the integral sign.

Exercise 7.10 This exercise deals with the concept of improper integrals and teaches students how to evaluate them.

Exercise 7.11 This exercise covers various miscellaneous integrals and teaches students how to solve problems using different methods.

Miscellaneous Exercise: This exercise contains a collection of problems that require students to apply the concepts they have learned in the previous exercises.


In summary, this chapter covers a wide range of topics related to integrals, and the exercises are designed to help students develop a deep understanding of the concepts and their applications.


Access NCERT Solutions for Class 12 Mathematics  Chapter 7- Integrals

Exercise 7.1

1. Find an antiderivative (or integral) of the following functions by the method of inspection. sin 2x

Ans:  We use the method of inspection as follows:

$ \dfrac{d}{dx}\left( \cos 2x \right)=-2\sin 2x\Rightarrow -\dfrac{1}{2}\dfrac{d}{dx}\left( \cos 2x \right) $

$\therefore \sin 2x=\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x \right)$

Thus, the anti-derivative of sin $2x$is $-\dfrac{1}{2}\cos 2x$.


2. Find an antiderivative (or integral) of the following functions by the method of inspection. cos 3x

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( \sin 3x \right)=3\cos 3x\Rightarrow \dfrac{1}{3}\dfrac{d}{dx}\left( \sin 3x \right)$

 $ \therefore \cos 3x=\dfrac{d}{dx}\left( \dfrac{1}{3}\sin 3x \right)$

        Thus, the anti - derivative of cos $3x$is $\dfrac{1}{3}\sin 3x$.


3. Find an antiderivative (or integral) of the following functions by the method of inspection. $\mathbf{{{e}^{2x}}}$

Ans: We use the method of inspection as follows:

$ \dfrac{d}{dx}\left( {{e}^{2x}} \right)\Rightarrow 2{{e}^{2x}}=\dfrac{1}{2}\dfrac{d}{dx}\left( {{e}^{2x}} \right) $

$ \therefore {{e}^{2x}}=\dfrac{d}{dx}\left( \dfrac{1}{2}{{e}^{2x}} \right) $

Thus, the anti-derivative of ${{e}^{2x}}$is $\dfrac{1}{2}{{e}^{2x}}$.


4. Find an antiderivative (or integral) of the following functions by the method of inspection.$\mathbf{{{\left( ax+b \right)}^{2}}}$

Ans: We use the method of inspection as follows:

  $  \dfrac{d}{dx}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}} $ 

$ \Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{1}{3a}\dfrac{d}{dx}{{\left( ax+b \right)}^{3}} $ 

 $  \therefore {{\left( ax+b \right)}^{2}}=\dfrac{d}{dx}\left( \dfrac{1}{3a}{{\left( ax+b \right)}^{3}} \right)$

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\dfrac{1}{3a}{{\left( ax+b \right)}^{3}}$.


5. Find an antiderivative (or integral) of the following functions by the method of inspection. sin $\mathbf{2x-4{{e}^{3x}}}$

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)=\left( \sin 2x-4{{e}^{3x}} \right)$

Thus, the anti-derivative of $\left( \sin 2x-4{{e}^{3x}} \right)$ is $\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)$.


6. $\int{\left( 4{{e}^{3x}}+1 \right)dx}$

Ans: 

$ \int{\left( 4{{e}^{3x}}+1 \right)dx} $

$ =4\int{{{e}^{3x}}dx}+\int{1}dx $

$ =4\left( \dfrac{{{e}^{3x}}}{3} \right)+x+C $ 

$=\dfrac{4}{3}{{e}^{3x}}+x+C$ 


7. $\int{{{x}^{2}}\left( 1-\dfrac{1}{{{x}^{2}}} \right)dx}$

Ans: $ \int{{{x}^{2}}\left( 1-\dfrac{1}{{{x}^{2}}} \right)dx} $

$=\int{\left( {{x}^{2}}-1 \right)dx} $

$ =\dfrac{{{x}^{3}}}{3}-x+C$


8. $\int{\left( a{{x}^{2}}+bx+c \right)dx}$

Ans:$ \int{\left( a{{x}^{2}}+bx+c \right)}dx $ 

$ =a\int{{{x}^{2}}dx+b\int{xdx+c\int{1.dx}}}$ 

$=a\left( \dfrac{{{x}^{3}}}{3} \right)+b\left( \dfrac{{{x}^{2}}}{2} \right)+cx+C $ 


9. $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx}$

Ans:  $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx} $

$ =2\int{{{x}^{2}}dx+\int{{{e}^{x}}dx}} $

$ =2\left( \dfrac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C$

$=\dfrac{2}{3}{{x}^{3}}+{{e}^{x}}+C $


10. $\int{{{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}dx}$

Ans:$ \int{{{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}}dx $

$ =\int{\left( x+\dfrac{1}{x}-2 \right)dx} $

$ =\int{xdx}+\int{\dfrac{1}{x}dx}-2\int{1.dx} $

$=\dfrac{{{x}^{2}}}{2}+\log \left| x \right|-2x+C $


11. $\int{\dfrac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

Ans: $\int{\dfrac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

\[ \int{\left( x+5-4{{x}^{-2}} \right)dx} \]

\[ =\int{xdx}+5\int{1.dx}-4\int{{{x}^{-2}}dx}\] 

\[=\dfrac{{{x}^{2}}}{2}+5x+\dfrac{4}{x}+C \]


12. $\int{\dfrac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

Ans:$\int{\dfrac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

$ =\int{\left( {{x}^{\dfrac{5}{2}}}+3{{x}^{\dfrac{1}{2}}}+4{{x}^{-\dfrac{1}{2}}} \right)dx} $

$ =\dfrac{{{x}^{\dfrac{7}{2}}}}{\dfrac{7}{2}}+\dfrac{3\left( {{x}^{\dfrac{3}{2}}} \right)}{\dfrac{3}{2}}+\dfrac{4\left( {{x}^{\dfrac{1}{2}}} \right)}{\dfrac{1}{2}}+C$

$ =\dfrac{2}{7}{{x}^{\dfrac{7}{2}}}+2{{x}^{\dfrac{3}{2}}}+8\sqrt{x}+C$


13. $\int{\dfrac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

Ans: $\int{\dfrac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

We obtain, on dividing:

$ =\int{\left( {{x}^{2}}+1 \right)dx} $

$ =\int{{{x}^{2}}dx}+\int{1.dx} $

$ =\dfrac{{{x}^{3}}}{3}+x+C$


14. $\int{\left( 1-x \right)}\sqrt{x}dx$

Ans: $\int{\left( 1-x \right)}\sqrt{x}dx$

 $=\int{\left( \sqrt{x}-{{x}^{\dfrac{3}{2}}} \right)dx} $

$ =\int{{{x}^{\dfrac{1}{2}}}dx-\int{{{x}^{\dfrac{3}{2}}}dx}} $ 

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}-\dfrac{2}{5}{{x}^{\dfrac{5}{2}}}+C $


15. $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

Ans: $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

$ =3\int{\left( 2{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+3{{x}^{\dfrac{1}{2}}} \right)} $

$=3\int{{{x}^{\dfrac{5}{2}}}dx+2\int{{{x}^{\dfrac{3}{2}}}dx+3\int{{{x}^{\dfrac{1}{2}}}}dx}} $

$=\dfrac{6}{7}{{x}^{\dfrac{7}{2}}}+\dfrac{4}{5}{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+C $


16. $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

Ans: $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

\[ =2\int{xdx-3\int{\cos xdx}+\int{{{e}^{x}}dx}} \]

\[=\dfrac{2{{x}^{2}}}{2}-3\left( \sin x \right)+{{e}^{x}}+C \]

\[ ={{x}^{2}}-3\sin x+{{e}^{x}}+C\]


17. $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

Ans: $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

=\[ 2\int{{{x}^{2}}dx-3\int{\sin xdx+5\int{{{x}^{\dfrac{1}{2}}}}dx}}\]

= \[ \dfrac{2{{x}^{3}}}{3}-3\left( -\cos x \right)+5\left(\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C\] 

=$ \dfrac{2}{3}{{x}^{3}}+3\cos x+\dfrac{10}{3}{{x}^{\dfrac{3}{2}}}+C $


18. $\int{\sec x\left( \sec x+\tan x \right)dx}$

Ans: $\int{\sec x\left( \sec x+\tan x \right)dx}$

 $ =\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx}$

$ =\int{{{\sec }^{2}}xdx+\int{\sec x\tan xdx}}$

$ =\tan x+\sec x+C $


19. $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

Ans: $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

$ =\int{\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}}dx} $

$=\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx}$

$=\int{{{\tan }^{2}}xdx}$

$ =\int{{{\sec }^{2}}xdx}-\int{1dx} $

$=\tan x-x+C $


20. $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

Ans: $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

$ =\int{\left( \dfrac{2}{{{\cos }^{2}}x}-\dfrac{3\sin x}{{{\cos }^{2}}x} \right)dx} $

$ =\int{2{{\sec }^{2}}xdx}-3\int{\tan x\sec xdx} $

$=2\tan x-3\sec x+C$


21. The anti – derivative of $\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)$ equals

  1. $\dfrac{1}{3}{{x}^{\dfrac{1}{3}}}+2{{x}^{\dfrac{1}{2}}}+C$

  2. $\dfrac{2}{3}{{x}^{\dfrac{2}{3}}}+\dfrac{1}{2}{{x}^{2}}+C$

  3. $\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}}+C$

  4. $\dfrac{3}{2}{{x}^{\dfrac{3}{2}}}+\dfrac{1}{2}{{x}^{\dfrac{1}{2}}}+C$

Ans: 

 $ \left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right) $

$=\int{{{x}^{\dfrac{1}{2}}}dx}+\int{{{x}^{-\dfrac{1}{2}}}}dx=\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{{{x}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C $ 

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}}+C $ 

 Thus, the correct answer is C.


22. If $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$ such that $f\left( 2 \right)=0$ then $f\left( x \right)$ is

  1. ${{x}^{4}}+\dfrac{1}{{{x}^{3}}}-\dfrac{129}{8}$

  2. ${{x}^{3}}+\dfrac{1}{{{x}^{4}}}+\dfrac{129}{8}$

  3. ${{x}^{4}}+\dfrac{1}{{{x}^{3}}}+\dfrac{129}{8}$

  4. ${{x}^{3}}+\dfrac{1}{{{x}^{4}}}-\dfrac{129}{8}$

Ans: Given, $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$ 

Anti-derivative of $4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)$

\[ \therefore f\left( x \right)=\int{4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)} \]

$ f\left( x \right)=4\int{{{x}^{3}}dx-3\int{\left( {{x}^{-4}} \right)}dx} $

$f\left( x \right)=4\left( \dfrac{{{x}^{4}}}{4} \right)-3\left( \dfrac{{{x}^{-3}}}{-3} \right)+C $

$ f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}+C$

Also, 

$f\left( 2 \right)=0 $

$\therefore f\left( 2 \right)={{\left( 2 \right)}^{4}}+\dfrac{1}{{{\left( 2 \right)}^{3}}}+C=0 $ 

$ \Rightarrow 16+\dfrac{1}{8}+C=0 $ 

$ \Rightarrow C=\dfrac{-129}{8} $

$\therefore f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}-\dfrac{129}{8} $

 Thus, the correct answer is A.


Exercise 7.2

1. $\dfrac{2x}{1+{{x}^{2}}}$

Ans: Substitute $1+{{x}^{2}}=t$

$\therefore 2xdx=dt $

$\Rightarrow \int{\dfrac{2x}{1+{{x}^{2}}}dx}=\int{\dfrac{1}{t}}dt$

$ =\log \left| t \right|+C$

$ =\log \left| 1+{{x}^{2}} \right|+C$

$ =\log \left( 1+{{x}^{2}} \right)+C $


2. $\dfrac{{{\left( \log x \right)}^{2}}}{x}$

Ans: Substitute $\log \left| x \right|=t$

$ \therefore \dfrac{1}{x}dx=dt $

$\Rightarrow \int{\dfrac{{{\left( \log \left| x \right| \right)}^{2}}}{x}}dx=\int{{{t}^{2}}dt} $

$=\dfrac{{{t}^{3}}}{3}+C$

$ =\dfrac{{{\left( \log \left| x \right| \right)}^{3}}}{3}+C $


3. ${\dfrac{1}{x+x\log x}$

Ans: $\dfrac{1}{x+x\log x}=\dfrac{1}{x\left( 1+\log x \right)}$

Substitute $1+\log x=t$

$ \therefore \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x\left( 1+\log x \right)}dx}=\int{\dfrac{1}{t}dt} $

$ =\log \left| t \right|+C$

$=\log \left| 1+\log x \right|+C $          


4. ${Sinx.\sin \left( \cos x \right)$

Ans: $\operatorname{Sin}x.\sin \left( \cos x \right)$

 $ Put,\cos x=t $ 

 $ \therefore -\sin xdx=dt $ 

 $ \Rightarrow \int{\sin x.\sin \left( \cos x \right)dx}=-\int{\sin tdt} $

 $ =-\left[ -\cos t \right]+C $ 

 $ =\cos t+C $ 

 $ =\cos \left( \cos x \right)+C $


5. $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)$

Ans: $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)=\dfrac{2\sin \left( ax+b \right)\cos \left( ax+b \right)}{2}=\dfrac{\sin 2\left( ax+b \right)}{2}$ 

Substitute $2\left( ax+b \right)=t$

$ \therefore 2adx=dt $

$ \Rightarrow \int{\dfrac{\sin 2\left( ax+b \right)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin t}{2a}}dt $

$ =\dfrac{1}{4a}\left[ -\cos t \right]+C $

$ =\dfrac{-1}{4a}\cos 2\left( ax+b \right)+C$


6. $\sqrt{ax+b}$

Ans: Substitute $ax+b=t$

$ \Rightarrow adx=dt $

$ \therefore dx=\dfrac{1}{a}dt $

 $ \Rightarrow \int{{{\left( ax+b \right)}^{\dfrac{1}{2}}}dx}=\dfrac{1}{a}\int{{{t}^{\dfrac{1}{2}}}}dt $ 

 $=\dfrac{1}{a}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{3}{2}} \right)+C=\dfrac{2}{3a}{{\left( ax+b \right)}^{\dfrac{3}{2}}}+C $


7. $x\sqrt{x+2}$

Ans: Substitute $x+2=t$

$\therefore dx=dt $

$ \Rightarrow \int{x\sqrt{x+2}}=\int{\left( t-2 \right)\sqrt{t}}dt$

$ =\int{\left( {{t}^{\dfrac{3}{2}}}-2{{t}^{\dfrac{1}{2}}} \right)}dt $

$ =\int{{{t}^{\dfrac{3}{2}}}dt-2\int{{{t}^{\dfrac{1}{2}}}}dt} $

$ =\dfrac{2}{5}{{t}^{\dfrac{5}{2}}}-\dfrac{4}{3}{{t}^{\dfrac{3}{2}}}+C $


8. $x\sqrt{1+2{{x}^{2}}}$

Ans: Substitute $1+2{{x}^{2}}=t$

$ \therefore 4xdx=dt $

$\Rightarrow \int{x\sqrt{1+2{{x}^{2}}}dx}=\int{\dfrac{\sqrt{t}dt}{4}} $

$ =\dfrac{1}{4}\int{{{t}^{\dfrac{1}{2}}}}dt $

$ =\dfrac{1}{4}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $

$=\dfrac{1}{6}{{\left( 1+2{{x}^{2}} \right)}^{\dfrac{3}{2}}}+C$


9. $\left( 4x+2 \right)\sqrt{{{x}^{2}}+x+1}$

Ans: Substitute ${{x}^{2}}+x+1=t$

$ \therefore \left( 2x+1 \right)dx=dt $

$ \int{\left( 4x+2 \right)\sqrt{{{x}^{2}}+x+1}}dx$

$ =\int{2\sqrt{t}dt}$

$=2\int{\sqrt{t}}dt $

$ =2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C=\dfrac{4}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{3}{2}}}+C $


10. $\dfrac{1}{x-\sqrt{x}}$

Ans: $\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)}$

Substitute $\left( \sqrt{x-1} \right)=t$

$\Rightarrow \int{\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx=\int{\dfrac{2}{t}dt}} $ 

$ =2\log \left| t \right|+C$ 

$=2\log \left| \sqrt{x}-1 \right|+C$


11. $\dfrac{x}{\sqrt{x+4}},x>0$

Ans: Substitute $x+4=t$

$ \therefore dx=dt $

$ \int{\dfrac{x}{\sqrt{x+4}}dx}=\int{\dfrac{\left( t-4 \right)}{\sqrt{t}}dt=\int{\left( \sqrt{t}-\dfrac{4}{\sqrt{t}} \right)}}dt $

 $ =\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-4\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C=\dfrac{2}{3}{{\left( t \right)}^{\dfrac{3}{2}}}-8{{\left( t \right)}^{\dfrac{1}{2}}}+C $

$ =\dfrac{2}{3}t.{{t}^{\dfrac{1}{2}}}-8{{t}^{\dfrac{1}{2}}}+C $

 $ =\dfrac{2}{3}{{t}^{\dfrac{1}{2}}}\left( t-12 \right)+C $

$=\dfrac{2}{3}\sqrt{x+4}\left( x-8 \right)+C $


12. ${{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}}$

Ans: Substitute ${{x}^{3}}-1=t$

$ \therefore 3{{x}^{2}}dx=dt $

$ \Rightarrow \int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}}dx=\int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{3}}.{{x}^{2}}dx}} $

$=\int{{{t}^{\dfrac{1}{3}}}\left( t+1 \right)\dfrac{dt}{3}=\dfrac{1}{3}\int{\left( {{t}^{\dfrac{4}{3}}}+{{t}^{\dfrac{1}{3}}} \right)dt}} $

$ =\dfrac{1}{3}\left[ \dfrac{3}{7}{{t}^{\dfrac{7}{3}}}+\dfrac{3}{4}{{t}^{\dfrac{4}{3}}} \right]+C $

$=\dfrac{1}{7}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{7}{3}}}+\dfrac{1}{4}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{4}{3}}}+C $


13. $\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}}$

Ans: Substitute $2+3{{x}^{3}}=t$

$ \therefore 9{{x}^{2}}dx=dt $

$ \Rightarrow \int{\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}}dx}=\dfrac{1}{9}\int{\dfrac{dt}{{{\left( t \right)}^{3}}}} $

$ =\dfrac{1}{9}\left[ \dfrac{{{t}^{-2}}}{-2} \right]+C $

$=\dfrac{-1}{18{{\left( 2+3{{x}^{3}} \right)}^{2}}}+C $ 


14. $\dfrac{1}{x{{\left( \log x \right)}^{m}}},x>0$

Ans: Substitute $\log x=t$

$ \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x{{\left( \log x \right)}^{m}}}dx}=\int{\dfrac{dt}{{{\left( t \right)}^{m}}}=\left( \dfrac{{{t}^{-m-1}}}{1-m} \right)}+C $

$ =\dfrac{{{\left( \log x \right)}^{1-m}}}{\left( 1-m \right)}+C $


15. $\dfrac{x}{9-4{{x}^{2}}}$

Ans: Substitute $9-4{{x}^{2}}=t$

$ \therefore -8xdx=dt $

$ \Rightarrow \int{\dfrac{x}{9-4{{x}^{2}}}dx=\dfrac{-1}{8}\int{\dfrac{1}{t}dt}} $

$=\dfrac{-1}{8}\log \left| t \right|+C $

$=\dfrac{-1}{8}\log \left| 9-4{{x}^{2}} \right|+C$


16. ${{e}^{2x+3}}$

Ans: Substitute $2x+3=t$

$\therefore 2dx=dt $

$ \Rightarrow \int{{{e}^{2x+3}}dx}=\dfrac{1}{2}\int{{{e}^{t}}dt} $

$ =\dfrac{1}{2}\left( {{e}^{t}} \right)+C $

$ =\dfrac{1}{2}{{e}^{\left( 2x+3 \right)}}+C $


17. $\dfrac{x}{{{e}^{{{x}^{2}}}}}$

Ans: Substitute ${{x}^{2}}=t$

$ \therefore 2xdx=dt$

$ \Rightarrow \int{\dfrac{x}{{{e}^{{{x}^{2}}}}}dx=\dfrac{1}{2}\int{\dfrac{1}{{{e}^{t}}}dt}} $

$=\dfrac{1}{2}\int{{{e}^{-t}}dt} $

$ =\dfrac{1}{2}\left( \dfrac{{{e}^{-t}}}{-1} \right)+C$ 

$=-\dfrac{1}{2}{{e}^{-{{x}^{2}}}}+C $

$ =\dfrac{-1}{2{{e}^{{{x}^{2}}}}}+C $


18. $\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$

Ans: Substitute ${{\tan }^{-1}}x=t$

$ \therefore \dfrac{1}{1+{{x}^{2}}}dx=dt $ 

$ \Rightarrow \int{\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=dt} $ 

 $  ={{e}^{t}}+C $ 

$ ={{e}^{{{\tan }^{-1}}x}}+C$ 


19. Solve the following: $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Ans: Given expression $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Let us substitute ${{e}^{2x}}+{{e}^{-2x}}=t$, we get

$\left( 2{{e}^{2x}}+2{{e}^{-2x}} \right)dx=dt$

$\Rightarrow 2\left( {{e}^{2x}}-{{e}^{-2x}} \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| t \right|+C$

Again substitute $t={{e}^{2x}}+{{e}^{-2x}}$, we get

$\therefore\int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C$


20.  Solve the following: ${{\tan }^{2}}\left( 2x-3 \right)$.

Ans: Given expression ${{\tan }^{2}}\left( 2x-3 \right)$.

We can apply the identity ${{\tan }^{2}}x={{\sec }^{2}}x-1$, we get

${{\tan }^{2}}\left( 2x-3 \right)={{\sec }^{2}}\left( 2x-3 \right)-1$

Substitute $2x-3=t$, we get

$2dx=dt$ 

Integration of given expression is 

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\int{{{\sec }^{2}}\left( 2x-3 \right)-1dx}$

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\int{{{\sec }^{2}}tdt-\int{1dx}}$

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\tan t-x+C$

Substitute $2x-3=t$

$\therefore \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\tan \left( 2x-3 \right)-x+C$


21.  Solve the following: ${{\sec }^{2}}\left( 7-4x \right)$.

Ans: Given expression ${{\sec }^{2}}\left( 7-4x \right)$.

Put $7-4x=t$, we get

$\therefore -4dx=dt$ 

Integration of given expression is

 $\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\int{{{\sec }^{2}}tdt}}$ 

$\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan t+C}$

Substitute $7-4x=t$, we get

$\therefore \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan \left( 7-4x \right)+C}$


22.  Solve the following: $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Put ${{\sin }^{-1}}x=t$, we get

$\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\int{tdt}}$

$\Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\dfrac{{{t}^{2}}}{2}+C}$ 

Substitute ${{\sin }^{-1}}x=t$, we get

$\therefore \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\dfrac{{{\left( {{\sin }^{-1}}x \right)}^{2}}}{2}+C}$


23.  Solve the following: $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Ans: Given expression is $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Given expression can be written as

 $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}=\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}$

Let $3\cos x+2\sin x=t$, we get

$\left( -3\sin x+2\cos x \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}}dx$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| t \right|+C$

Substitute $3\cos x+2\sin x=t$

$\therefore \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| 2\sin x+3\cos x \right|+C$


24. Solve the following: $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Ans: Given expression $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Given expression can be written as 

$\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}=\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}$

Let $\left( 1-\tan x \right)=t$, we get

$-{{\sec }^{2}}xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}}dx$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{-dt}{{{t}^{2}}}}$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=-\int{{{t}^{-2}}dt}$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\dfrac{1}{t}+C$

Substitute $\left( 1-\tan x \right)=t$, 

$\therefore \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\dfrac{1}{\left( 1-\tan x \right)}+C$


25. Solve the following: $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Ans: Given expression is $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Let $\sqrt{x}=t$, we get

$\dfrac{1}{2\sqrt{x}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\int{\cos tdt}$

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin t+C$

Substitute $\sqrt{x}=t$

$\therefore \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin \sqrt{x}+C$


26. Solve the following: $\sqrt{\sin 2x}\cos 2x$.

Ans: Given expression is $\sqrt{\sin 2x}\cos 2x$.

Let $\sin 2x=t$, we get

$2\cos 2xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\int{\sqrt{t}dt}}$ 

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C}$

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+C}$

Substitute $\sin 2x=t$

$\therefore \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}+C}$


27. Solve the following: $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Ans: Given expression $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Let $1+\sin x=t$ 

$\therefore \cos xdx=dt$ 

Integration of given expression is  

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\int{\dfrac{dt}{\sqrt{t}}}$ 

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C$

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{t}+C$

Substitute $1+\sin x=t$,

$\therefore \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{1+\sin x}+C$


28.  Solve the following: $\cot x\log \sin x$.

Ans: Given expression $\cot x\log \sin x$.

Let $\log \sin x=t$, we get

$\dfrac{1}{\sin x}\cos xdx=dt$ 

$\Rightarrow \cot xdx=dt$ 

Integration of given expression is

$\int{\cot x\log \sin xdx=\int{tdt}}$ 

$\Rightarrow \int{\cot x\log \sin xdx=\dfrac{{{t}^{2}}}{2}+C}$

Substitute $\log \sin x=t$,

$\therefore \int{\cot x\log \sin xdx=\dfrac{1}{2}{{\left( \log \sin x \right)}^{2}}+C}$


29.  Solve the following: $\dfrac{\sin x}{1+\cos x}$.

Ans: Given expression $\dfrac{\sin x}{1+\cos x}$.

Let $1+\cos x=t$ 

$\therefore -\sin xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=\int{-\dfrac{dt}{t}}}$ 

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| t \right|+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| 1+\cos x \right|+C}$


30.  Solve the following: $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Ans: Given expression $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Let $1+\cos x=t$ 

$\therefore -\sin xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\int{-\dfrac{dt}{{{t}^{2}}}}}$ 

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=-\int{{{t}^{-2}}dt}}$

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{t}+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{1+\cos x}+C}$

.

31.Solve the following: $\dfrac{1}{1+\cot x}$.

Ans: Given expression $\dfrac{1}{1+\cot x}$.

Let  $I=\int{\dfrac{1}{1+\cot x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1+\cot x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1+\dfrac{\cos x}{\sin x}}}dx$

$\Rightarrow I=\int{\dfrac{\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)+\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)}{\sin x+\cos x}+\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}}dx$

Let $\sin x+\cos x=t$ 

$\therefore \left( \cos x-\sin x \right)dx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\sin x+\cos x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \sin x+\cos x \right|+C$


32.  Solve the following: $\dfrac{1}{1-\tan x}$.

Ans: Given expression $\dfrac{1}{1-\tan x}$.

Let  $I=\int{\dfrac{1}{1-\tan x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1-\tan x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1-\dfrac{\sin x}{\cos x}}}dx$

$\Rightarrow I=\int{\dfrac{\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)+\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)}{\cos x-\sin x}+\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}}dx$

Let $\cos x-\sin x=t$ 

$\therefore \left( -\sin x-\cos x \right)dx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\cos x-\sin x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \cos x-\sin x \right|+C$


33. Solve the following: $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Ans: Given expression $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Let $I=\int{\dfrac{\sqrt{\tan x}}{\sin x\cos x}dx}$ 

Multiply and divide by $\cos x$, we get

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\sin x\cos x\times \cos x}dx}$

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\tan x\times {{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{\sqrt{\tan x}}dx}$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\int{\dfrac{dt}{\sqrt{t}}dx}$

$\Rightarrow I=2\sqrt{t}+C$

Substitute $\tan x=t$,

$\therefore I=2\sqrt{\tan x}+C$


34.  Solve the following: $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Let $1+\log x=t$ 

$\therefore \dfrac{1}{x}dx=dt$ 

Integration of given expression is

$\int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\int{{{t}^{2}}dt}}$ 

$\Rightarrow \int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\dfrac{{{t}^{3}}}{3}+C}$

Substitute $1+\log x=t$

$\therefore \int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\dfrac{{{\left( 1+\log x \right)}^{3}}}{3}+C}$


35. Solve the following: $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Given expression can be written as

$\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( \dfrac{x+1}{x} \right){{\left( x+\log x \right)}^{2}}$

$\Rightarrow \dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( 1+\dfrac{1}{x} \right){{\left( x+\log x \right)}^{2}}$

Let $x+\log x=t$ 

$\therefore \left( 1+\dfrac{1}{x} \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\int{{{t}^{2}}dt}}$ 

$\Rightarrow \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\dfrac{{{t}^{3}}}{3}+C}$

Substitute $x+\log x=t$

$\therefore \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\dfrac{1}{3}{{\left( x+\log x \right)}^{3}}+C}$


36. Solve the following: $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$. 

Ans: Given expression $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$.

Let ${{x}^{4}}=t$,

$\therefore 4{{x}^{3}}dx=dt$ 

Integration of given expression is

 $\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\dfrac{\sin \left( {{\tan }^{-1}}t \right)}{1+{{t}^{2}}}dt}}$ ………(1)

Let ${{\tan }^{-1}}t=u$ 

$\therefore \dfrac{1}{1+{{t}^{2}}}dt=du$ 

Substitute in eq. (1), we get

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\sin udu}}$

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\left( -\cos u \right)+C}$

Substitute ${{\tan }^{-1}}t=u$,

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}t \right)+C}$

Substitute ${{x}^{4}}=t$,

$\therefore \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}{{x}^{4}} \right)+C}$


37. $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}}dx$ equals 

  1. ${{10}^{x}}-{{x}^{10}}+C$ 

  2. ${{10}^{x}}+{{x}^{10}}+C$

  3. ${{\left( {{10}^{x}}-{{x}^{10}} \right)}^{-1}}+C$

  4. $\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Ans: Given expression $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}$.

Let ${{x}^{10}}+{{10}^{x}}=t$,

$\therefore \left( 10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10 \right)dx=dt$  

Integration of given expression is

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\int{\dfrac{dt}{t}}$ 

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log t+C$

Substitute ${{x}^{10}}+{{10}^{x}}=t$,

$\therefore \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Therefore, option D is the correct answer.


38. $\int{\dfrac{dx}{{{\sin }^{2}}{{\cos }^{2}}x}}$ equals

  1. $\tan x+\cot x+C$ 

  2. $\tan x-\cot x+C$

  3. $\tan x\cot x+C$

  4. $\tan x-\cot 2x+C$

Ans: Given expression $\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$.

Let  $I=\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$ 

$I=\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}+\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{1}{{{\cos }^{2}}x}dx}+\int{\dfrac{1}{{{\sin }^{2}}x}dx}$

$\Rightarrow I=\int{{{\sec }^{2}}xdx}+\int{cose{{c}^{2}}xdx}$

$\Rightarrow I=\tan x-\cot x+C$

Therefore, option B is the correct answer.


Exercise 7.3

1.  Solve the following: ${{\sin }^{2}}\left( 2x+5 \right)$.

Ans: Given expression ${{\sin }^{2}}\left( 2x+5 \right)$.

Given expression can be written as  

${{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos 2\left( 2x+5 \right)}{2}$

$\Rightarrow {{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos \left( 4x+10 \right)}{2}$ 

Integration of given expression is

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\int{\dfrac{1-\cos \left( 4x+10 \right)}{2}dx}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}\int{1dx-\dfrac{1}{2}\int{\cos \left( 4x+10 \right)dx}}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{2}\left( \dfrac{\sin \left( 4x+10 \right)}{4} \right)+C$

$\therefore \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{8}\sin \left( 4x+10 \right)+C$


2. Solve the following: $\sin 3x\cos 4x$.

Ans: Given expression $\sin 3x\cos 4x$.

Using the identity $\sin A\cos B=\dfrac{1}{2}\left\{ \sin \left( A+B \right)+\sin \left( A-B \right) \right\}$ given expression can be written as

$\sin 3x\cos 4x=\dfrac{1}{2}\left\{ \sin \left( 3x+4x \right)+\sin \left( 3x-4x \right) \right\}$

Integration of above expression is

$\Rightarrow \int{\sin 3x\cos 4x}dx=\int{\dfrac{1}{2}\left\{ \sin 7x+\sin \left( -x \right) \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\left\{ \sin 7x+\sin x \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\sin 7xdx+\dfrac{1}{2}\int{\sin x}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\left( \dfrac{-\cos 7x}{7} \right)-\dfrac{1}{2}\left( -\cos x \right)+C$

$\therefore \int{\sin 3x\cos 4x}dx=\dfrac{-\cos 7x}{14}+\dfrac{\cos x}{2}+C$


3.  Solve the following: $\cos 2x\cos 4x\cos 6x$.

Ans: Given expression $\cos 2x\cos 4x\cos 6x$.

Using the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$ given expression can be written as

$\cos 2x\left( \cos 4x\cos 6x \right)=\cos 2x\dfrac{1}{2}\left\{ \cos \left( 4x+6x \right)+\cos \left( 4x-6x \right) \right\}$

Integration of the above expression is

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\cos 2x}\left[ \dfrac{1}{2}\left\{ \cos 10x+\cos \left( -2x \right) \right\} \right]dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\left[ \dfrac{1}{2}\left\{ \cos 2x\cos 10x+\cos 2x\cos \left( -2x \right) \right\} \right]}dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \cos 2x\cos 10x+{{\cos }^{2}}2x \right\} \right]}dx$

Again applying the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$, we get

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \dfrac{1}{2}\cos \left( 2x+10x \right)+\cos \left( 2x-10x \right)+\left( \dfrac{1+\cos 4x}{2} \right) \right\} \right]}dx$$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{4}\int{\left[ \cos 12x+\cos 8x+\cos 4x \right]}dx$

$\therefore \int{\cos 2x\cos 4x\cos 6x}=\dfrac{1}{4}\left[ \dfrac{\sin 12x}{12}+\dfrac{\sin 8x}{8}+\dfrac{\sin 4x}{4} \right]+C$


4. Solve the following: ${{\sin }^{3}}\left( 2x+1 \right)$.

Ans: Given expression ${{\sin }^{3}}\left( 2x+1 \right)$. 

Let $I=\int{{{\sin }^{3}}\left( 2x+1 \right)dx}$

$\Rightarrow I=\int{{{\sin }^{2}}\left( 2x+1 \right)\sin \left( 2x+1 \right)dx}$ 

$\Rightarrow I=\int{\left( 1-{{\cos }^{2}}\left( 2x+1 \right) \right)\sin \left( 2x+1 \right)dx}$

Let $\cos \left( 2x+1 \right)=t$ 

$\therefore -2\sin \left( 2x+1 \right)dx=dt$ 

Integration becomes

$\Rightarrow I=-\dfrac{1}{2}\int{\left( 1-{{t}^{2}} \right)dt}$

$\Rightarrow I=-\dfrac{1}{2}\left( t-\dfrac{{{t}^{3}}}{3} \right)+C$

Substitute $\cos \left( 2x+1 \right)=t$,

$\Rightarrow I=-\dfrac{1}{2}\left( \cos \left( 2x+1 \right)-\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3} \right)+C$

$\therefore \int{{{\sin }^{3}}\left( 2x+1 \right)dx}=\dfrac{-\cos \left( 2x+1 \right)}{2}+\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3}+C$


5. Solve the following: ${{\sin }^{3}}x{{\cos }^{3}}x$.

Ans: Given expression ${{\sin }^{3}}x{{\cos }^{3}}x$.

Let $I=\int{{{\sin }^{3}}x{{\cos }^{3}}x}dx$ 

$\Rightarrow I=\int{{{\sin }^{2}}x\sin x{{\cos }^{3}}x}dx$

$\Rightarrow I=\int{{{\cos }^{3}}x\left( 1-{{\cos }^{2}}x \right)\sin x}dx$

Let $\cos x=t$ 

$\therefore -\sin dx=dt$ 

Integration becomes

$\Rightarrow I=-\int{{{t}^{3}}\left( 1-{{t}^{2}} \right)}dt$

$\Rightarrow I=-\int{\left( {{t}^{3}}-{{t}^{5}} \right)}dt$

$\Rightarrow I=-\left[ \dfrac{{{t}^{4}}}{4}-\dfrac{{{t}^{6}}}{6} \right]+C$

Substitute $\cos x=t$,

$\Rightarrow I=-\left[ \dfrac{{{\cos }^{4}}x}{4}-\dfrac{{{\cos }^{6}}x}{6} \right]+C$

$\therefore \int{{{\sin }^{3}}x{{\cos }^{3}}x}dx=\dfrac{{{\cos }^{6}}x}{6}-\dfrac{{{\cos }^{4}}x}{4}+C$


6. Solve the following: $\sin x\sin 2x\sin 3x$.  

Ans: Given expression $\sin x\sin 2x\sin 3x$.

Using the identity $\sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

$\Rightarrow \sin x\sin 2x\sin 3x=\sin x.\dfrac{1}{2}\cos \left( 2x-3x \right)-\cos \left( 2x+3x \right)$

Integration of given expression is

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos \left( -x \right)-\sin x\cos \left( 5x \right) \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos x-\sin x\cos 5x \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\dfrac{\sin 2x}{2}dx-\dfrac{1}{2}\int{\left( \sin x\cos 5x \right)}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{4}\left( \dfrac{-\cos 2x}{2} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{1}{2}\left( \sin \left( x+5x \right)+\sin \left( x-5x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin \left( -4x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin 4x \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{8}\left[ \dfrac{-\cos 6x}{3}+\dfrac{\cos 4x}{4} \right]+C$

$\therefore \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{8}\left[ \dfrac{\cos 6x}{3}-\dfrac{\cos 4x}{4}-\cos 2x \right]+C$


7.  Solve the following: $\sin 4x\sin 8x$. 

Ans: Given expression $\sin 4x\sin 8x$.

Using the identity $\sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

$\Rightarrow \sin 4x\sin 8x=\dfrac{1}{2}\cos \left( 4x-8x \right)-\cos \left( 4x+8x \right)$

Integration of given expression is

$\Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos \left( -4x \right)-\cos \left( 12x \right) \right)dx}$

$\Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx}$

$\therefore \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\left[ \dfrac{\sin 4x}{4}-\dfrac{\sin 12x}{12} \right]+C$


8.  Solve the following: $\dfrac{1-\cos x}{1+\cos x}$. 

Ans: Given expression $\dfrac{1-\cos x}{1+\cos x}$.

Using the identities $2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ given expression can be written as 

$\Rightarrow \dfrac{1-\cos x}{1+\cos x}=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2}$

Integration of given expression is 

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\tan }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\sec }^{2}}\dfrac{x}{2}-1 \right]dx}$

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\left[ \dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}-x \right]+C$

$\therefore \int{\dfrac{1-\cos x}{1+\cos x}dx}=2\tan \dfrac{x}{2}-x+C$


9.  Solve the following: $\dfrac{\cos x}{1+\cos x}$. 

Ans: Given expression $\dfrac{\cos x}{1+\cos x}$.

Using the identity $\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ given expression can be written as 

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-\dfrac{{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}} \right]$

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right]$

Integration of given expression is 

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\sec }^{2}}\dfrac{x}{2}+1 \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 2-{{\sec }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\left[ 2x-\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}} \right]+C$

$\therefore \int{\dfrac{\cos x}{1+\cos x}dx}=x-\tan \dfrac{x}{2}+C$


10.  Solve the following: ${{\sin }^{4}}x$. 

Ans: Given expression ${{\sin }^{4}}x$.

Given expression can be written as ${{\sin }^{4}}x={{\sin }^{2}}x{{\sin }^{2}}x$

$\Rightarrow {{\sin }^{4}}x=\left( \dfrac{1-\cos 2x}{2} \right)\left( \dfrac{1-\cos 2x}{2} \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}{{\left( 1-\cos 2x \right)}^{2}}$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 4x}{2}-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right)$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\int{\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right)}dx$

$\Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\left[ \dfrac{3}{2}x+\dfrac{\sin 4x}{8}-\sin 2x \right]+C$

$\therefore \int{{{\sin }^{4}}x}dx=\dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{1}{4}\sin 2x+C$

 

11.  Solve the following: ${{\cos }^{4}}2x$.

Ans: Given expression ${{\cos }^{4}}2x$.

Given expression can be written as  

${{\cos }^{4}}2x={{\left( {{\cos }^{2}}2x \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x={{\left( \dfrac{1+\cos 4x}{2} \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}4x+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 8x}{2}+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right)$

Integration of given expression is 

$\Rightarrow \int{{{\cos }^{4}}2xdx}=\int{\left( \dfrac{3}{8}+\dfrac{\cos 8x}{8}+\dfrac{\cos 4x}{2} \right)dx}$

$\therefore \int{{{\cos }^{4}}2xdx}=\dfrac{3}{8}x+\dfrac{\sin 8x}{64}+\dfrac{\sin 4x}{8}+C$


12. Solve the following: $\dfrac{{{\sin }^{2}}x}{1+\cos x}$.

Ans: Given expression $\dfrac{{{\sin }^{2}}x}{1+\cos x}$.

By applying the identity $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$, given expression can be written as 

  $\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{{{\left( 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=2{{\sin }^{2}}\dfrac{x}{2}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=1-\cos x$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=\int{1dx-\int{\cos xdx}}$

$\therefore \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=x-\sin x+C$


13.  Solve the following: $\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$. 

Ans: Given expression $\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$.

We can apply the identity $\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ , we get

 $\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{-2\sin \dfrac{2x+2\alpha }{2}\sin \dfrac{2x-2\alpha }{2}}{-2\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \dfrac{2\left( x+\alpha  \right)}{2}\sin \dfrac{2\left( x-\alpha  \right)}{2}}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \left( x+\alpha  \right)\sin \left( x-\alpha  \right)}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\left[ 2\sin \dfrac{x+\alpha }{2}\cos \dfrac{x+\alpha }{2} \right]\left[ 2\sin \dfrac{x-\alpha }{2}\cos \dfrac{x-\alpha }{2} \right]}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=4\cos \dfrac{x+\alpha }{2}\cos \dfrac{x-\alpha }{2}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos \dfrac{x+\alpha }{2}+\dfrac{x-\alpha }{2}+\cos \dfrac{x+\alpha }{2}-\dfrac{x-\alpha }{2} \right]$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos x+\cos \alpha  \right]$

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\int{\left[ \cos x+\cos \alpha  \right]}dx$

$\therefore \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\left[ \sin x+x\cos \alpha  \right]+C$


14. Solve the following: $\dfrac{\cos x-\sin x}{1+\sin 2x}$. 

Ans: Given expression $\dfrac{\cos x-\sin x}{1+\sin 2x}$.

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Given expression can be written as  

$\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}$ .

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

 $\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}$

$\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}$

Let $\sin x+\cos x=t$ 

$\therefore \left( \cos x-\sin x \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}}$ 

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{dt}{{{t}^{2}}}}}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{{{t}^{-2}}dt}}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-{{t}^{-1}}+C}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{t}+C}$

Substitute $\sin x+\cos x=t$,

$\therefore \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{\sin x+\cos x}+C}$


15.  Solve the following: ${{\tan }^{3}}2x\sec 2x$.

Ans: Given expression ${{\tan }^{3}}2x\sec 2x$.

Given expression can be written as

${{\tan }^{3}}2x\sec 2x={{\tan }^{2}}2x\tan 2x\sec 2x$

$\Rightarrow {{\tan }^{3}}2x\sec 2x=\left( {{\sec }^{2}}2x-1 \right)\tan 2x\sec 2x$

$\Rightarrow {{\tan }^{3}}2x\sec 2x={{\sec }^{2}}2x\tan 2x\sec 2x-\tan 2x\sec 2x$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\int{\tan 2x\sec 2xdx}$

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\dfrac{\sec 2x}{2}+C$

Let $\sec 2x=t$ 

$\therefore 2\sec 2x\tan 2xdx=dt$ 

Above integral becomes

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{1}{2}\int{{{t}^{2}}dt}-\dfrac{\sec 2x}{2}+C$

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{t}^{3}}}{6}-\dfrac{\sec 2x}{2}+C$

Substitute $\sec 2x=t$,

$\therefore \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{\left( \sec 2x \right)}^{3}}}{6}-\dfrac{\sec 2x}{2}+C$


16.  Solve the following: ${{\tan }^{4}}x$. 

Ans: Given expression ${{\tan }^{4}}x$.

Given expression can be written as 

$\Rightarrow {{\tan }^{4}}x={{\tan }^{2}}x{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x=\left( {{\sec }^{2}}x-1 \right){{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-\left( {{\sec }^{2}}x-1 \right)$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1 \right)}dx$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x \right)}dx-\int{{{\sec }^{2}}xdx+\int{1dx}}$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{\sec }^{2}}x{{\tan }^{2}}x}dx-\tan x+x+C$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{t}^{2}}}dt-\tan x+x+C$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\dfrac{{{t}^{3}}}{3}-\tan x+x+C$

Substitute $\tan x=t$,

$\therefore \int{{{\tan }^{4}}xdx}=\dfrac{1}{3}{{\tan }^{3}}x-\tan x+x+C$


17. Solve the following: $\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

Ans: Given expression $\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

Given expression can be written as 

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{{{\sin }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}+\dfrac{{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\sin x}{{{\cos }^{2}}x}+\dfrac{\cos x}{{{\sin }^{2}}x}$

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\tan x\sec x+\cot xcosecx$

Integration of given expression is 

$\Rightarrow \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\tan x\sec xdx}+\int{\cot x\operatorname{cosecx}dx}$

$\therefore \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\sec x-cosecx+C$


18.  Solve the following: $\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

Ans: Given expression $\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

By applying the identity $\cos 2x=1-2{{\sin }^{2}}x$, we get

 $\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{\cos 2x+1-\cos 2x}{{{\cos }^{2}}x}$

$\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$

$\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\sec }^{2}}x$

Integration of given expression is 

$\Rightarrow \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\int{{{\sec }^{2}}xdx}$

$\therefore \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\tan x+C$


19.  Solve the following: $\dfrac{1}{\sin x{{\cos }^{3}}x}$.

Ans: Given expression $\dfrac{1}{\sin x{{\cos }^{3}}x}$.

We can apply the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x}{\sin x{{\cos }^{3}}x}+\dfrac{{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{\sin x}{{{\cos }^{3}}x}+\dfrac{1}{\sin x\cos x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\cos }^{2}}x}{\dfrac{\sin x\cos x}{{{\cos }^{2}}x}}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\sec }^{2}}x}{\tan x}$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{\tan x{{\sec }^{2}}x}dx+\int{\dfrac{{{\sec }^{2}}x}{\tan x}}dx$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{t}dt+\int{\dfrac{1}{\operatorname{t}}}dt$

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{{{t}^{2}}}{2}+\log \left| t \right|+C$

Substitute $\tan x=t$,

$\therefore \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{1}{2}{{\tan }^{2}}x+\log \left| \tan x \right|+C$


20. Solve the following: $\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$. 

Ans: Given expression $\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$.

Given expression can be written as 

$\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $2\sin x\cos x=\sin 2x$, we get

 $\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+\sin 2x}$

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx}$

Let $1+\sin 2x=t$ 

$\therefore 2\cos 2xdx=dt$ 

Integration becomes

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| t \right|+C$

Substitute $1+\sin 2x=t$

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| 1+\sin 2x \right|+C$

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| {{\left( \cos x+\sin x \right)}^{2}} \right|+C$

$\therefore \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\log \left| \left( \cos x+\sin x \right) \right|+C$


21.  Solve the following: ${{\sin }^{-1}}\left( \cos x \right)$. 

Ans: Given expression ${{\sin }^{-1}}\left( \cos x \right)$.

Let $\cos x=t$ 

$\therefore \sin x=\sqrt{1-{{t}^{2}}}$ 

$\Rightarrow -\sin xdx=dt$

$\Rightarrow dx=-\dfrac{dt}{\sin x}$

$\Rightarrow dx=-\dfrac{dt}{\sqrt{1-{{t}^{2}}}}$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{{{\sin }^{-1}}t\left( \dfrac{-dt}{\sqrt{1-{{t}^{2}}}} \right)}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\int{\left( \dfrac{{{\sin }^{-1}}t}{\sqrt{1-{{t}^{2}}}} \right)dt}$

Let ${{\sin }^{-1}}t=u$ 

$\Rightarrow \dfrac{1}{\sqrt{1-{{t}^{2}}}}dt=du$ 

Integration becomes

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{4du}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{u}^{2}}}{2}+C$

Substitute ${{\sin }^{-1}}t=u$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\left( {{\sin }^{-1}}t \right)}^{2}}}{2}+C$

Substitute $\cos x=t$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\left[ {{\sin }^{-1}}\left( \cos x \right) \right]}^{2}}}{2}+C$ ……..(1)

We know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ 

$\therefore {{\sin }^{-1}}\left( \cos x \right)=\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \cos x \right)=\left( \dfrac{\pi }{2}-x \right)$ 

Substitute in eq. (1), we get

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{-{{\left( \dfrac{\pi }{2}-x \right)}^{2}}}{2}+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{1}{2}\left( \dfrac{{{\pi }^{2}}}{2}+{{x}^{2}}-\pi x \right)+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\pi x+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+\left( C-\dfrac{{{\pi }^{2}}}{4} \right)$

$\therefore \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+{{C}_{1}}$


22. Solve the following: $\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$. 

Ans: Given expression $\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( x-b \right)\cos \left( x-a \right)-\cos \left( x-b \right)\sin \left( x-a \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]dx}$

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ -\log \left| \cos \left( x-b \right) \right|+\log \cos \left( x-a \right) \right]}$

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \log \left| \dfrac{\cos \left( x-a \right)}{\cos \left( x-b \right)} \right| \right]}+C$


23. Solve the following: $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ is equal to

  1. $\tan x+\cot x+C$ 

  2. $\tan x+cosecx+C$ 

  3. $-\tan x+\cot x+C$ 

  4. $\tan x+\sec x+C$ 

Ans: Given expression $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.

Given expression can be written as

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}-\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{{{\sec }^{2}}xdx}-\int{cose{{c}^{2}}xdx}$

$\therefore \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\tan x+\cot x+C$

Therefore, option A is the correct answer.


24. Solve the following: $\int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$ equals

  1. $-\cot \left( e{{x}^{x}} \right)+C$ 

  2. $\tan \left( x{{e}^{x}} \right)+C$ 

  3. $\tan \left( {{e}^{x}} \right)+C$

  4. $\cot \left( {{e}^{x}} \right)+C$

Ans: Given expression $\int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$.

Let ${{e}^{x}}x=t$ 

$\therefore \left( {{e}^{x}}.x+{{e}^{x}}.1 \right)dx=dt$ 

$\Rightarrow {{e}^{x}}\left( x+1 \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{\dfrac{dt}{{{\cos }^{2}}t}}$

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{{{\sec }^{2}}t}dt$

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan t+C$

Substitute ${{e}^{x}}x=t$,

$\therefore \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan \left( {{e}^{x}}x \right)+C$

Therefore, option B is the correct answer.


Exercise 7.4

1. Solve the following: $\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}$.

Ans: Given expression $\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx=\int{\dfrac{dt}{{{t}^{2}}+1}}}$ 

We know that $\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x}$ 

$\Rightarrow \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx={{\tan }^{-1}}t+C}$

Substitute ${{x}^{3}}=t$,

$\therefore \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx={{\tan }^{-1}}\left( {{x}^{3}} \right)+C}$


2. Solve the following: $\dfrac{1}{\sqrt{1+4{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{1+4{{x}^{2}}}}$.

Let $2x=t$ 

$\therefore 2dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{1+{{t}^{2}}}}}}$

We know that $\int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\log \left| t+\sqrt{{{t}^{2}}+1} \right|}+C$

Substitute $2x=t$,

$\therefore \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\log \left| 2x+\sqrt{4{{x}^{2}}+1} \right|}+C$


3. Solve the following: $\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}$.

Ans: Given expression $\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}$.

Let $2-x=t$ 

$\therefore -dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\int{\dfrac{1}{\sqrt{{{t}^{2}}+1}}dt}}$ 

We know that $\int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\log \left| t+\sqrt{{{t}^{2}}+1} \right|+C}$

Substitute $2-x=t$,

$\Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\log \left| \left( 2-x \right)+\sqrt{{{\left( 2-x \right)}^{2}}+1} \right|+C}$

$\therefore \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=\log \left| \dfrac{1}{\left( 2-x \right)+\sqrt{{{x}^{2}}-4x+5}} \right|+C}$


4. Solve the following: $\dfrac{1}{\sqrt{9-25{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{9-25{{x}^{2}}}}$.

Let $5x=t$ 

$\therefore 5dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{9-{{t}^{2}}}}dt}}$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{{{3}^{2}}-{{t}^{2}}}}dt}}$

$\Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{t}{3} \right)+C}$

Substitute $5x=t$,

$\therefore \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C}$


5.  Solve the following: $\dfrac{3x}{1+2{{x}^{4}}}$.

Ans: Given expression $\dfrac{3x}{1+2{{x}^{4}}}$.

Let $\sqrt{2}{{x}^{2}}=t$ 

$\therefore 2\sqrt{2}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}\int{\dfrac{dt}{1+{{t}^{2}}}}$ 

$\Rightarrow \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}{{\tan }^{-1}}t+C$

Substitute $\sqrt{2}{{x}^{2}}=t$,

$\therefore \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}{{\tan }^{-1}}\left( \sqrt{2}{{x}^{2}} \right)+C$


6.  Solve the following: $\dfrac{{{x}^{2}}}{1-{{x}^{6}}}$.

Ans: Given expression $\dfrac{{{x}^{2}}}{1-{{x}^{6}}}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\int{\dfrac{dt}{1-{{t}^{2}}}}}$  

$\Rightarrow \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}\log \left| \dfrac{1+t}{1-t} \right| \right]+C}$

Substitute ${{x}^{3}}=t$,

$\therefore \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}\log \left| \dfrac{1+{{x}^{3}}}{1-{{x}^{3}}} \right| \right]+C}$


7.  Solve the following: $\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}$.

Ans: Given expression $\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}$.

Given expression can be written as

$\Rightarrow \dfrac{x-1}{\sqrt{{{x}^{2}}-1}}=\dfrac{x}{\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\sqrt{{{x}^{2}}-1}}$

Integration of given expression is

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}-\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}dx}$

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$

Let ${{x}^{2}}-1=t$ 

$\therefore 2xdx=dt$ 

Integration becomes

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}dt}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\left( 2{{t}^{\dfrac{1}{2}}} \right)-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\sqrt{t}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$

Substitute ${{x}^{2}}-1=t$

$\Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\sqrt{{{x}^{2}}-1}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$


8.  Solve the following: $\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}$.

Ans: Given expression $\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\int{\dfrac{dt}{\sqrt{{{t}^{2}}+{{\left( {{a}^{3}} \right)}^{2}}}}}}$ 

$\Rightarrow \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\log \left| t+\sqrt{{{t}^{2}}+{{a}^{6}}} \right|+C}$

Substitute ${{x}^{3}}=t$,

$\therefore \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\log \left| {{x}^{3}}+\sqrt{{{x}^{6}}+{{a}^{6}}} \right|+C}$


9. Solve the following: $\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}$.

Ans: Given expression $\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}$.

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\int{\dfrac{dt}{\sqrt{{{t}^{2}}+{{2}^{2}}}}}}$ 

$\Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\log \left| t+\sqrt{{{\operatorname{t}}^{2}}+4} \right|}+C$

Substitute $\tan x=t$,

$\therefore \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\log \left| \tan x+\sqrt{{{\tan }^{2}}x+4} \right|}+C$


10.  Solve the following: $\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}$.

Ans: Given expression $\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}$.

Given expression can be written as 

$\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}=\dfrac{1}{\sqrt{{{\left( x+1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}}$

Let $x+1=t$ 

$\therefore dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}+1}}dt}$

$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| t+\sqrt{{{t}^{2}}+1} \right|+C$

Substitute $x+1=t$,

$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|+C$

$\therefore \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| \left( x+1 \right)+\sqrt{{{x}^{2}}+2x+2} \right|+C$


11.  Solve the following: $\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}$.

Ans: Given expression $\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}$.

Given expression can be written as 

$\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}=\dfrac{1}{\sqrt{{{\left( 3x+1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}}$

Let $3x+1=t$ 

$\therefore 3dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{{{t}^{2}}+{{2}^{2}}}}dt}}$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{t}{2} \right) \right]+C}$

Substitute $3x+1=t$,

$\therefore \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{3x+1}{2} \right) \right]+C}$


12.  Solve the following: $\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{7-\left( {{x}^{2}}+6x+9-9 \right)}}$

$\Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{16-\left( {{x}^{2}}+6x+9 \right)}}$

$\Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}}$

$\Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( x+3 \right)}^{2}}}}$

Let $x+3=t$ 

$\therefore dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( t \right)}^{2}}}}dt}$

$\Rightarrow \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{t}{4} \right)+C$

Substitute $x+3=t$,

$\therefore \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C$


13.  Solve the following: $\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}$.

Ans: Given expression $\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}$.

Given expression can be written as 

$\Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-3x+2}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4}+2}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{\left( x-\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}$

Let $x-\dfrac{3}{2}=t$ 

$\therefore dx=dt$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}dx}$

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\log \left| t+\sqrt{{{t}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}} \right|+C$

Substitute $x-\dfrac{3}{2}=t$,

$\therefore \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\log \left| \left( x-\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}-3x+2} \right|+C$


14.  Solve the following: $\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{8-\left( {{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4} \right)}}$ 

$\Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{8-\left( {{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4} \right)}}$

$\Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{\dfrac{41}{4}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}}$

Let $x-\dfrac{3}{2}=t$ 

$\therefore dx=dt$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{\left( \dfrac{\sqrt{41}}{2} \right)}^{2}}-{{\left( t \right)}^{2}}}}dt}$

$\Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{t}{\dfrac{\sqrt{41}}{2}} \right)+C$

Substitute $x-\dfrac{3}{2}=t$

$\Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}} \right)+C$

$\therefore \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{2x-3}{\sqrt{41}} \right)+C$


15.  Solve the following: $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$.

Ans: Given expression $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+\dfrac{{{\left( a+b \right)}^{2}}}{4}-\dfrac{{{\left( a+b \right)}^{2}}}{4}+ab}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-\dfrac{{{\left( a+b \right)}^{2}}}{4}}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx}$

Let $x-\left( \dfrac{a+b}{2} \right)=t$ 

$\therefore dx=dt$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx}$

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| t+\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}} \right|+C$

Substitute $x-\left( \dfrac{a+b}{2} \right)=t$,

$\therefore \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| x-\left( \dfrac{a+b}{2} \right)+\sqrt{\left( x-a \right)\left( x-b \right)} \right|+C$


16. $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$

Ans: $(x-a)(x-b)={{x}^{2}}-(a+b)x+ab$

${{x}^{2}}-(a+b)x+ab={{x}^{2}}-(a+b)x+\dfrac{{{(a+b)}^{2}}}{4}-\dfrac{{{(a+b)}^{2}}}{4}+ab$

Simplifying, 

$={{\left[ x-\left( \dfrac{a+b}{2} \right) \right]}^{2}}-\dfrac{{{(a-b)}^{2}}}{4}$

$\Rightarrow \int{\dfrac{1}{\sqrt{(x-a)(x-b)}}}dx=\int{\dfrac{1}{\sqrt{{{\left\{ x-\left( \dfrac{a+b}{2} \right) \right\}}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}}dx$

Consider $\text{ }x-\left( \dfrac{a+b}{2} \right)=t\therefore dx=dt$

$\int{\dfrac{1}{\sqrt{{{\left\{ x-\left( \dfrac{a+b}{2} \right) \right\}}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}}dt\text{ }$

Using the logarithm formula of integration, 

$=log\left| t+\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}} \right|+C$

Substitute the value of t, 

$=\log \left| \left\{ x-\left( \dfrac{a+b}{2} \right) \right\}+\sqrt{(x-a)(x-b)} \right|+\text{C}$


17. $\dfrac{4x+1}{\sqrt{2{{x}^{2}}+x-3}}$

Ans:Consider $4x+1=A\dfrac{d}{dx}\left( 2{{x}^{2}}+x-3 \right)+B$

Simplifying, 

$\Rightarrow 4x+1=A(4x+1)+B$

$\Rightarrow 4x+1=4Ax+A+B$

We obtain the below values by equating the coefficients of $\text{x}$ and the constant term on both sides. 

$4~\text{A}=4\Rightarrow \text{A}=1$

$\text{A}+\text{B}=1\Rightarrow \text{B}=0$

Consider $2{{x}^{2}}+x-3=t$

$\therefore (4x+1)dx=dt$

$\Rightarrow \int{\dfrac{4x+1}{\sqrt{2{{x}^{2}}+x-3}}}dx=\int{\dfrac{1}{\sqrt{t}}}dt$

Using the power rule of integration,

$=2\sqrt{t}+C$

Substitute the value of t,

$=2\sqrt{2{{x}^{2}}+x-3}+C$


18. $\dfrac{x+2}{\sqrt{{{x}^{2}}-1}}$

 Ans:Consider  $\text{x}+2=A\dfrac{d}{dx}\left( {{x}^{2}}-1 \right)+B$

$\Rightarrow x+2=A(2x)+B......\left( 1 \right)$

We obtain the below values by equating the coefficients of x  and    the constant term on both sides. 

$2A=1\Rightarrow A=\dfrac{1}{2}$

$B=2$

From (1), we get

$(x+2)=\dfrac{1}{2}(2x)+2$

$\int{\dfrac{x+2}{\sqrt{{{x}^{2}}-1}}}dx=\int{\dfrac{\dfrac{1}{2}(2x)+2}{\sqrt{{{x}^{2}}-1}}}dx$

$=\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx+\int{\dfrac{2}{\sqrt{{{x}^{2}}-1}}}dx$

$\text{In }\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx\text{ let }{{x}^{2}}-1=t\Rightarrow 2xdx=dt$

$\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}\text{ }$

Integrating using the power rule

$=\dfrac{1}{2}[2\sqrt{t}]$

                   Simplifying,

$=\sqrt{t}\text{ }$

Substitute the value of t,

$=\sqrt{{{x}^{2}}-1}$

Then, $\int{\dfrac{2}{\sqrt{{{x}^{2}}-1}}}dx=2\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}}dx=2\log \left| x+\sqrt{{{x}^{2}}-1} \right|$

From equation (2), we get

$\int{\dfrac{x+2}{\sqrt{{{x}^{2}}-1}}}dx=\sqrt{{{x}^{2}}-1}+2\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C$


19. $\dfrac{5x-2}{1+2x+3{{x}^{2}}}$

Ans: Let $5x-2=A\dfrac{d}{dx}\left( 1+2x+3{{x}^{2}} \right)+B\text{ }$

$\Rightarrow \text{ }5\text{ }x-2=A\left( 2+6\text{ }x \right)+B......\left( 1 \right)$

We obtain the below values by equating the coefficients of x  and            

the constant term on both sides. 

$5=6A\Rightarrow A=\dfrac{5}{6}$

$2A+B=-2\Rightarrow B=-\dfrac{11}{3}$

Substitute the above values in (1)

$\therefore 5x-2=\dfrac{5}{6}(2+6x)+\left( -\dfrac{11}{3} \right)$

$\Rightarrow\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\int{\dfrac{\dfrac{5}{6}(2+6x)-\dfrac{11}{3}}{1+2x+3{{x}^{2}}}}dx$

$=\dfrac{5}{6}\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx-\dfrac{11}{3}\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx$

Consider ${{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx$

$\therefore\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\dfrac{5}{6}{{I}_{1}}-\dfrac{11}{3}{{I}_{2}}...\left( 1 \right)$

${{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx$

Put $1+2x+3{{x}^{2}}=t$

$\Rightarrow (2+6x)dx=dt$

$\therefore {{I}_{1}}=\int{\dfrac{dt}{t}}$

Using the logarithm formula of integration,

${{I}_{1}}=\log |t|\text{ }$

Substitute the value of t,

${{I}_{1}}=\log \left| 1+2x+3{{x}^{2}} \right|...\left( 2 \right)$

Then, 

${{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx$

$1+2x+3{{x}^{2}}$ can be rewritten as $1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right)$

Thus,

$1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right)$

By completing square method, 

$=1+3\left( {{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{1}{9} \right)$

$=1+3{{\left( x+\dfrac{1}{3} \right)}^{2}}-\dfrac{1}{3}$

Simplifying,

$=\dfrac{2}{3}+3{{\left( x+\dfrac{1}{3} \right)}^{2}}$

$=3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+\dfrac{2}{9} \right]$

$=3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right]$

Therefore ${{I}_{2}}$can be rewritten as , 

${{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{\left. {{\left[ x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right]}}dx$

$=\dfrac{1}{3}\left[ \dfrac{3}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right) \right]$

Simplifying,

$=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right)...\left( 3 \right)$

We obtain the below values by substituting equations (2) and (3) in equation (1)

$\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\dfrac{5}{6}\left[ \log \left| 1+2x+3{{x}^{2}} \right| \right]-\dfrac{11}{3}\left[ \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right) \right]+C$

Simplifying,

$=\dfrac{5}{6}\log \left| 1+2x+3{{x}^{2}} \right|-\dfrac{11}{3\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right)+C$


20. $\dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}}$

 Ans: Consider $6x+7=A\dfrac{d}{dx}\left( {{x}^{2}}-9x+20 \right)+B$

Differentiating, 

$\Rightarrow 6\text{ }x+7=A\left( 2\text{ }x-9 \right)+B$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$2~\text{A}=6\Rightarrow \text{A}=3$

$-9~\text{A}+\text{B}=7\Rightarrow \text{B}=34$

$\therefore 6\text{x}+7=3(2\text{x}-9)+34$

$\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=\int{\dfrac{3(2x-9)+34}{\sqrt{{{x}^{2}}-9x+20}}}dx$

$=3\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx+34\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Consider  ${{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

$\therefore\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=3{{I}_{1}}+34{{I}_{2}}\,\,\,\,\,\,\,\,....(1)$

${{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Put  ${{x}^{2}}-9x+20=t$

$\Rightarrow (2x-9)dx=dt$

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}$

Integrating using the power rule

${{I}_{1}}=2\sqrt{t}$

Substitute the value of t,

${{I}_{1}}=2\sqrt{{{x}^{2}}-9x+20}\,\,\,.....(2)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Consider 

${{x}^{2}}-9x+20$

By completing square methods, 

$={{x}^{2}}-9x+20+\dfrac{81}{4}-\dfrac{81}{4}$

$={{\left( x-\dfrac{9}{2} \right)}^{2}}-\dfrac{1}{4}$

$={{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}$

$\Rightarrow {{I}_{2}}=\int{\dfrac{1}{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}\,\,dx$

${{I}_{2}}=\log \left| \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right|.....\left( 3 \right)$

We obtain the below values by substituting equations (2) and (3) in (1), $\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}dx=3\left[ 2\sqrt{{{x}^{2}}-9x+20} \right]+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+\text{C}$

Simplifying,

$=6\sqrt{{{x}^{2}}-9x+20}+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+C$


21. $\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}$

Ans:Consider, $\text{x}+2=A\dfrac{d}{dx}\left( 4x-{{x}^{2}} \right)+B$

$\Rightarrow x+2=A(4-2x)+B$

 We obtain the below values by equating the coefficients of x  and the  

constant term on both sides.

$-2A=1\Rightarrow A=-\dfrac{1}{2}$

 $4A+B=2\Rightarrow B=4$

$\Rightarrow (x+2)=-\dfrac{1}{2}(4-2x)+4$

$\therefore\int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=\int{\dfrac{-\dfrac{1}{2}(4-2x)+4}{\sqrt{4x-{{x}^{2}}}}}dx$    $=-\dfrac{1}{2}\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx+4\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx$

Let ${{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx$ and ${{I}_{2}}\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx$

$\therefore \int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=-\dfrac{1}{2}{{I}_{1}}\text{ and }+4{{I}_{2}}\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx$

 Let $4x-{{x}^{2}}=t$

$\Rightarrow (4-2x)dx=dt$

$\Rightarrow{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{4x-{{x}^{2}}}...\left( 2 \right)$

(Using the logarithm formula of integration,)

${{I}_{2}}=\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx$

Integrating using the power rule, 

$\Rightarrow 4x-{{x}^{2}}=-\left( -4x+{{x}^{2}} \right)$

By completing square methods, 

$=\left( -4x+{{x}^{2}}+4-4 \right)$

$=4-{{(x-2)}^{2}}$

$={{(2)}^{2}}-{{(x-2)}^{2}}$

$\therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(2)}^{2}}-{{(x-2)}^{2}}}}}dx={{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)...\left( 3 \right)$

   Using equations (2) and (3) in (1), to obtain 

$\int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=-\dfrac{1}{2}\left( 2\sqrt{4x-{{x}^{2}}} \right)+4{{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)+C$

$=-\sqrt{4x-{{x}^{2}}}+4{{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)+C$


22.$\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}$

Ans:$\int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}\int{\dfrac{2(x+2)}{\sqrt{{{x}^{2}}+2x+3}}}dx$

Simplifying,

$=\dfrac{1}{2}\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+2x+3}}}dx$

$=\dfrac{1}{2}\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx+\dfrac{1}{2}\int{\dfrac{2}{\sqrt{{{x}^{2}}+2x+3}}}dx$

$=\dfrac{1}{2}\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx+\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx$

Let ${{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx$

$\therefore \int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}{{I}_{1}}+{{I}_{2}}\,\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx$

Put, ${{x}^{2}}+2x+3=t$

Integrating using the power rule, 

$\Rightarrow (2\text{x}+2)\text{dx}=\text{dt}\,\,\,\,\,\,$

$\,{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+2x+3}..\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx.$

By completing square methods,

$\Rightarrow {{x}^{2}}+2x+3={{x}^{2}}+2x+1+2={{(x+1)}^{2}}+{{(\sqrt{2})}^{2}}$

$\therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(x+1)}^{2}}+{{(\sqrt{2})}^{2}}}}}dx=\log \left| (x+1)+\sqrt{{{x}^{2}}+2x+3} \right|...\left( 3 \right)$

Using equations (2) and (3) in (1), to obtain                             

$\therefore\int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\sqrt{{{x}^{2}}+2x+3}+\log \left| (x+1)+\sqrt{{{x}^{2}}+2x+3} \right|\,+C$


23. $\dfrac{x+3}{{{x}^{2}}-2x-5}$

Ans:  Consider $(x+3)=A\dfrac{d}{dx}\left( {{x}^{2}}-2x-5 \right)+B$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

We obtain the below values by equating the coefficients of x  and the constant term on both sides.

$2A=1\Rightarrow A=\dfrac{1}{2}$

$-2A+B=3\Rightarrow B=4$

$\therefore (x+3)=\dfrac{1}{2}(2x-2)+4$

$\Rightarrow \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\int{\dfrac{\dfrac{1}{2}(2x-2)+4}{{{x}^{2}}-2x-5}}dx$

$=\dfrac{1}{2}\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx+4\int{\dfrac{1}{{{x}^{2}}-2x-5}}~\text{d}x$

Consider${{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx$

$\therefore \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\dfrac{1}{2}{{I}_{1}}+4{{I}_{2}}...\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx$

Put ${{x}^{2}}-2x-5=\text{t}$

$\Rightarrow (2x-2)dx=dt$

Using the logarithm formula of integration,

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}-2x-5 \right|\,\,\,....\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx$

$=\int{\dfrac{1}{\left( {{x}^{2}}-2x+1 \right)-6}}dx$

$=\int{\dfrac{1}{{{(x-1)}^{2}}+{{(\sqrt{6})}^{2}}}}dx$

$=\dfrac{1}{2\sqrt{6}}\log \left( \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right)....\left( 3 \right)$

We obtain the below values by substituting (2) and (3) in (1), 

$\int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\dfrac{1}{2}\log \left| {{x}^{2}}-2x-5 \right|+\dfrac{4}{2\sqrt{6}}\log \left| \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+C$

$=\dfrac{1}{2}\log \left| {{x}^{2}}-2x-5 \right|+\dfrac{2}{\sqrt{6}}\log \left| \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+C$


24. $\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}$

Ans:  $5x+3=A\dfrac{a}{dx}\left( {{x}^{2}}+4x+10 \right)+B$

$\Rightarrow 5x+3=A(2x+4)+B$

Equating the coefficients of $\text{x}$and constant term, we get

$2A=5\Rightarrow A=\dfrac{5}{2}$

$4A+B=3\Rightarrow B=-7$

$\therefore 5x+3=\dfrac{5}{2}(2x+4)-7$

$\Rightarrow \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\int{\dfrac{\dfrac{5}{2}(2x+4)-7}{\sqrt{{{x}^{2}}+4x+10}}}dx$

$=\dfrac{5}{2}\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx-7\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx$

Let ${{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx$

$\therefore \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}{{I}_{1}}-7{{I}_{2}}\,\,\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx$

Consider ${{x}^{2}}+4x+10=\text{t}\therefore (2x+4)dx=dt$

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+4x+10}\,\,\,\,\,\,\,......\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx$

$=\int{\dfrac{1}{\sqrt{\left( {{x}^{2}}+4x+4 \right)+6}}}dx=\int{\dfrac{1}{{{(x+2)}^{2}}+{{(\sqrt{6})}^{2}}}}dx$

$=\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|...\left( 3 \right)$

We obtain the below values by using equations (2) and (3) in (1). $\int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}\left[ 2\sqrt{{{x}^{2}}+4x+10} \right]-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$$=5\sqrt{{{x}^{2}}+4x+10}-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$


25. $\int{\dfrac{dx}{{{x}^{2}}+2x+2}}$equals 

  1. $x{{\tan }^{-1}}\left( x+1 \right)+C$

  2. ${{\tan }^{-1}}\left( x+1 \right)+C$

  3. $\left( x+1 \right){{\tan }^{-1}}x+C$

  4. ${{\tan }^{-1}}x+C$

Ans: $\int{\dfrac{dx}{{{x}^{2}}+2x+2}}=\int{\dfrac{dx}{\left( {{x}^{2}}+2x+1 \right)+1}}$

$=\int{\dfrac{1}{{{(x+1)}^{2}}+{{(1)}^{2}}}}dx$

$=\left[ {{\tan }^{-1}}(x+1) \right]+C$

Hence, the right response is is B.


26. $\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$ equals 

  1. $\dfrac{1}{9}{{\sin }^{-1}}\left( \dfrac{9x-8}{8} \right)+C$

  2. $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{8x-9}{9} \right)+C$

  3. $\dfrac{1}{3}{{\sin }^{-1}}\left( \dfrac{9x-8}{8} \right)+C$

  4. $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{9x-8}{9} \right)+C$

Ans:

             $\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$

         $=\int{\dfrac{1}{\sqrt{-4\left( {{x}^{2}}-\dfrac{9}{4}x \right)}}}dx$

    By completing square methods,

     $=\int{\dfrac{1}{-4\left( {{x}^{2}}-\dfrac{9}{4}x+\dfrac{81}{64}-\dfrac{81}{64} \right)}}dx$

$=\int{\dfrac{1}{\sqrt{-4\left[ {{\left( x-\dfrac{9}{8} \right)}^{2}}-{{\left( \dfrac{9}{8} \right)}^{2}} \right]}}}dx$

$=\dfrac{1}{2}\int{\dfrac{1}{{{\left( \dfrac{9}{8} \right)}^{2}}-{{\left( x-\dfrac{9}{8} \right)}^{2}}}}dx$

$=\dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{x-\dfrac{9}{8}}{\dfrac{9}{8}} \right) \right]+C\quad \left( \int{\dfrac{dy}{\sqrt{{{a}^{2}}-{{y}^{2}}}}}={{\sin }^{-1}}\dfrac{y}{a}+C \right)$

   Simplifying,

 $=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{8x-9}{9} \right)+C$

Hence, the right response is B.


Exercise 7.5:

1. $\dfrac{x}{\left( x+1 \right)\left( x+2 \right)}$

Ans:   Let $\dfrac{x}{(x+1)(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$\Rightarrow x=A(x+2)+B(x+1)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$\text{A}+\text{B}=1$

$2~\text{A}+\text{B}=0$

On solving, we get

$\text{A}=-1\text{ and B}=2$

$\therefore \dfrac{x}{(x+1)(x+2)}=\dfrac{-1}{(x+1)}+\dfrac{2}{(x+2)}$

$\Rightarrow \int{\dfrac{x}{(x+1)(x+2)}}dx=\int{\dfrac{-1}{(x+1)}}+\dfrac{2}{(x+2)}dx$

Using the logarithm formula of integration,

$=-\log |x+1|+2\log |x+2|+C$

$=\log {{(x+2)}^{2}}-\log |x+1|+C$

Simplifying,

$=\log \dfrac{{{(x+2)}^{2}}}{(x+1)}+C$


2. $\dfrac{1}{{{x}^{2}}-9}$

Ans:  Let $\dfrac{1}{(x+3)(x-3)}=\dfrac{A}{(x+3)}+\dfrac{B}{(x-3)}$

$1=A\left( x-3 \right)+B\left( x+3 \right)$

Equating the coefficients of $\text{x}$and constant term, we get

$A+B=0$ 

$1=-3A+3B$

$A=-\dfrac{1}{6}\text{ and }B=\dfrac{1}{6}$

$\therefore \dfrac{1}{(x+3)(x-3)}=\dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)}$

$\Rightarrow \int{\dfrac{1}{\left( {{x}^{2}}-9 \right)}}dx=\int{\left( \dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)} \right)}dx$

Using the logarithm formula of integration,

$=-\dfrac{1}{6}\log |x+3|+\dfrac{1}{6}\log |x-3|+C=\dfrac{1}{6}\log \dfrac{|(x-3)|}{|(x+3)|}+C$


3. $\dfrac{3x-1}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans: Let $\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\quad \ldots \left( 1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$\text{A}+\text{B}+\text{C}=0$

$-5~\text{A}-4~\text{B}-3\text{C}=3$

$6~\text{A}+3~\text{B}+2\text{C}=-1$

Solving these equations, to obtain 

$\text{A}=1,~\text{B}=-5,\text{ and C}=4$

$\therefore \dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)}$

$\Rightarrow \int{\dfrac{3x-1}{(x-1)(x-2)(x-3)}}dx=\int{\left\{ \dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)} \right\}}dx$

Using the logarithm formula of integration,

$=\log |x-1|-5\log |x-2|+4\log |x-3|+C$


4. $\dfrac{x}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans:Let $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$x=A\left( x-2 \right)\left( x-3 \right)+B\left( x-1 \right)\left( x-3 \right)+C\left( x-1 \right)\left( x-2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0\text{ }$

$-5\text{ }A-4\text{ }B-3\text{ }C=1$

$6\text{ }A+4\text{ }B+2\text{ }C=0$

Solving these equations, to obtain 

$A=\dfrac{1}{2},B=2\text{ and }C=\dfrac{3}{2}$

$\therefore \dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)}$

$\Rightarrow \int{\dfrac{x}{(x-1)(x-2)(x-3)}}dx=\int{\left\{ \dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)} \right\}}dx$

Using the logarithm formula of integration,

$=\dfrac{1}{2}\log |x-1|-2\log |x-2|+\dfrac{3}{2}\log |x-3|+C$


5. $\dfrac{2x}{{{x}^{2}}+3x+2}$

 Ans:$\dfrac{2x}{{{x}^{2}}+3x+2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$2\text{ }x=A\left( x+2 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B=2$

$2\text{ }A+B=0$

Solving these equations, we get

$A=-2\text{ and }B=4$

$\therefore \dfrac{2x}{(x+1)(x+2)}=\dfrac{-2}{(x+1)}+\dfrac{4}{(x+2)}$

$\Rightarrow \int{\dfrac{2x}{(x+1)(x+2)}}dx=\int{\left\{ \dfrac{4}{(x+2)}-\dfrac{2}{(x+1)} \right\}}dx$

Using the logarithm formula of integration,

$=4\log |x+2|-2\log |x+1|+C$


6. $\dfrac{1-{{x}^{2}}}{x\left( 1-2x \right)}$

Ans:  It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $\left( 1-{{x}^{2}} \right)$ by $x(1-2x)$ to obtain, $\dfrac{1-{{x}^{2}}}{x(1+2x)}=\dfrac{1}{2}+\dfrac{1}{2}\left( \dfrac{2-x}{x(1-2x)} \right)$

Let $\dfrac{2-x}{x(1-2x)}=\dfrac{A}{x}+\dfrac{B}{(1-2x)}...\left( 1 \right)$

$\Rightarrow (2-x)=A(1-2x)+Bx$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

 $-2~\text{A}+\text{B}=-1$ and, $A=2$

Solving these equations, to obtain $A=2\text{ and }B=3$

$\therefore \dfrac{2-x}{x(1-2x)}=\dfrac{2}{x}+\dfrac{3}{1-2x}$

Substituting in equation (1), we get

$\dfrac{1-{{x}^{2}}}{x(1+2)}=\dfrac{1}{2}+\dfrac{1}{2}\left\{ \dfrac{2}{x}+\dfrac{3}{(1-2x)} \right\}$

$\Rightarrow \int{\dfrac{1-{{x}^{2}}}{x(1+2)}}dx=\int \dfrac{1}{2}+\dfrac{1}{2}\text{(}\dfrac{2}{x}+\dfrac{3}{(1-2x)})dx$

Using the power rule and logarithm formula of integration,

$=\dfrac{x}{2}+\log |x|+\dfrac{3}{2(-2)}\log |1-2x|+C=\dfrac{x}{2}+\log |x|-\dfrac{3}{4}\log |1-2x|+C$


7. $\dfrac{x}{\left( {{x}^{2}}+1 \right)\left( x-1 \right)}$

Ans: Let $\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{Ax+B}{\left( {{x}^{2}}+1 \right)}+\dfrac{C}{(x-1)}...\left( 1 \right)$

$x=(Ax+B)(x-1)+C\left( {{x}^{2}}+1 \right)$

$x=A{{x}^{2}}-Ax+Bx-B+C{{x}^{2}}+C$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0\text{ }$

$-A+B=1$

$-B+C=0$

On solving these equations, to obtain  $A=-\dfrac{1}{2},B=\dfrac{1}{2},and\,\,C=\dfrac{1}{2}$

From equation (1), to obtain  

$\therefore \dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{\left( -\dfrac{1}{2}x+\dfrac{1}{2} \right)}{{{x}^{2}}+1}+\dfrac{\dfrac{1}{2}}{(x-1)}$

$\Rightarrow \int{\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}}=-\dfrac{1}{2}\int{\dfrac{x}{{{x}^{2}}+1}}dx+\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}+1}}dx+\dfrac{1}{2}\int{\dfrac{1}{x-1}}dx$

$=-\dfrac{1}{4}\int{\dfrac{2x}{{{x}^{2}}+1}}dx+\dfrac{1}{2}{{\tan }^{-1}}x+\dfrac{1}{2}\log |x-1|+C$

Consider $\int{\dfrac{2x}{{{x}^{2}}+1}}dx,lel\left( {{x}^{2}}+1 \right)=t\Rightarrow 2xdx=dt$

$\Rightarrow \int{\dfrac{2x}{{{x}^{2}}+1}}dx=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}+1 \right|$

$\therefore \int{\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}}=-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+\dfrac{1}{2}\log |x-1|+C$


$=\dfrac{1}{2}\log |x-1|-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+C$


8. .$\dfrac{x}{{{\left( x-1 \right)}^{2}}\left( x+2 \right)}$

Ans: $\dfrac{x}{{{(x-1)}^{2}}(x+2)}$

$\text{ Let }\dfrac{x}{{{(x-1)}^{2}}(x+2)}=\dfrac{A}{(x-1)}+\dfrac{B}{{{(x-1)}^{2}}}+\dfrac{C}{(x+2)}$

$x=A(x-1)(x+2)+B(x+2)+C{{(x-1)}^{2}}$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0$

$A+B-2\text{ }C=1$

On solving, to obtain

$A=\dfrac{2}{9}\text{ and }C=\dfrac{-2}{9}$

$B=\dfrac{1}{3}$

$\therefore \dfrac{x}{{{(x-1)}^{2}}(x+2)}=\dfrac{2}{9(x-1)}+\dfrac{1}{3{{(x-1)}^{2}}}-\dfrac{2}{9(x\mid 2)}$

$\Rightarrow \int{\dfrac{x}{{{(x-1)}^{2}}(x+2)}}dx=\dfrac{2}{9}\int{\dfrac{1}{(x-1)}}dx+\dfrac{1}{3}\int{\dfrac{1}{{{(x-1)}^{2}}}}dx-\dfrac{2}{9}\int{\dfrac{1}{(x-2)}}dx$

Using the power rule and logarithm formula of integration,

$=\dfrac{2}{9}\log |x-1|+\dfrac{1}{3}\left( \dfrac{-1}{x-1} \right)-\dfrac{2}{9}\log |x+2|+C$

Simplifying,

$=\dfrac{2}{9}\log \left| \dfrac{x-1}{x+2} \right|-\dfrac{1}{3(x-1)}+C$


9. $\dfrac{3x+5}{{{x}^{3}}-{{x}^{2}}-x+1}$

Ans:   $\dfrac{3x+5}{{{x}^{3}}-{{x}^{2}}-x+1}=\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}$

let  $\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{A}{(x+1)}+\dfrac{B}{{{(x-1)}^{2}}}+\dfrac{C}{(x+1)}$

$3x+5=A(x-1)(x+1)+B(x+1)+{{(x-1)}^{2}}$

$3x+5=A{{(x-1)}^{2}}+B(x+1)+C\left( {{x}^{2}}+1-2x \right)\quad \ldots (1)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and  

the constant term on both sides.

$A+C=0\text{ }$

$B-2\text{ }C=3\text{ }$

$-A+B+C=5$

On solving, to obtain   $B=4\,\,A=-\dfrac{1}{2}\,\,and\,\,C=\dfrac{1}{2}$

$\therefore \dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{-1}{2(x-1)}+\dfrac{4}{{{(x-1)}^{2}}}+\dfrac{1}{2(x+1)}$

$\Rightarrow \int{\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}}dx=-\dfrac{1}{2}\int{\dfrac{1}{x-1}}dx+4\int{\dfrac{1}{{{(x-1)}^{2}}}}dx+\dfrac{1}{2}\int{\dfrac{1}{(x+1)}}dx$

Using the power rule and logarithm formula of integration,

$=-\dfrac{1}{2}\log |x-1|+4\left( \dfrac{-1}{x-1} \right)+\dfrac{1}{2}\log |x+1|+C$

$=\dfrac{1}{2}\log \left| \dfrac{x+1}{x-1} \right|-\dfrac{4}{(x-1)}+C$


10. $\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)\left( 2x+3 \right)}$

Ans: $\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)(2x+3)}=\dfrac{2x-3}{(x+1)(x-1)(2x+3)}$

Let $\dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{C}{(2x+3)}$

$\Rightarrow (2x-3)=A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$

$\Rightarrow (2x-3)-A\left( 2{{x}^{2}}+x-3 \right)+B\left( 2{{x}^{2}}+5x-3 \right)+C\left( {{x}^{2}}-1 \right)$

$\Rightarrow (2x-3)=(2A+2B+C){{x}^{2}}+(A+5B)x+(-3A+3B-C)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$2~\text{A}+2~\text{B}+\text{C}=0$

$\text{A}+5~\text{B}=2$

$-3~\text{A}+3~\text{B}-\text{C}=-3$

On solving, to obtain $B=-\dfrac{1}{10},A=\dfrac{5}{2},\text{ and C}=-\dfrac{24}{5}$

$\therefore \dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{5}{2(x+1)}-\dfrac{1}{10(x-1)}-\dfrac{24}{5(2x+3)}$

$\Rightarrow \int{\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)(2x+3)}}dx=\dfrac{5}{2}\int{\dfrac{1}{(x+1)}}dx-\dfrac{1}{10}\int{\dfrac{1}{x-1}}dx-\dfrac{24}{5}\int{\dfrac{1}{(2x+3)}}dx$

Using the logarithm formula of integration,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{24}{5\times 2}\log |2x+3|$

Simplifying,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{12}{5}\log |2x+3|+C$


11. $\dfrac{5x}{\left( x+1 \right)\left( {{x}^{2}}-4 \right)}$

Ans: $\dfrac{5x}{(x+1)\left( {{x}^{2}}-4 \right)}=\dfrac{5x}{(x+1)(x+2)(x-2)}$

let $\dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-2)}$

$5\text{ }x=A\left( x+2 \right)\left( x-2 \right)+B\left( x+1 \right)\left( x-2 \right)+C\left( x+1 \right)\left( x+2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0$

$-B+3C=5\text{ and, }-4A-2B+2C=0$

On solving, to obtain

$A=\dfrac{5}{3},B=-\dfrac{5}{2},\text{ and }C=\dfrac{5}{6}$

$\therefore \dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{5}{3(x+1)}+-\dfrac{5}{2(x+2)}+\dfrac{5}{6(x-2)}$

$\Rightarrow \int{\dfrac{5x}{(x+1)\left( {{x}^{2}}-4 \right)}}dx=\dfrac{5}{3}\int{\dfrac{1}{(x+1)}}dx-\dfrac{5}{2}\int{\dfrac{1}{(x+2)}}dx+\dfrac{5}{6}\int{\dfrac{1}{(x-2)}}dx$

Using the logarithm formula of integration,

$=\dfrac{5}{3}\log |x+1|-\dfrac{5}{2}\log |x+2|+\dfrac{5}{6}\log |x-2|+C$


12. $\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-1}$

 Ans:  On dividing $\left( {{x}^{3}}+x+1 \right)\,\,\,by\,\,{{x}^{2}}-1,$we get

$\dfrac{{{x}^{3}}+x+1}{{{x}^{2}}-1}=x+\dfrac{2x+1}{{{x}^{2}}-1}$

$\text{ Let }\dfrac{2x+1}{{{x}^{2}}-1}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}$

$2\text{ }x+1=A\left( x-1 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$A+B=2$

$-A+B=1$

On solving, to obtain  

$A=\dfrac{1}{2}\text{ and }B=\dfrac{3}{2}\text{ }$

$\therefore \dfrac{{{x}^{3}}+x+1}{{{x}^{2}}-1}=x+\dfrac{1}{2(x+1)}+\dfrac{3}{2(x-1)}\text{ }$

$\Rightarrow \int{\dfrac{{{x}^{3}}+x+1}{{{x}^{2}}+1}}dx=\int{x}dx+\dfrac{1}{2}\int{\dfrac{1}{(x+1)}}dx+\dfrac{3}{2}\int{\dfrac{1}{(x-1)}}dx$

$=\dfrac{{{x}^{2}}}{2}+\log |x+1|+\dfrac{3}{2}\log |x-1|+C$


13. $\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}$

Ans: Let $\dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(1-x)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}$

$2=A\left( 1+{{x}^{2}} \right)+(Bx+C)(1-x)$

$2=A+A{{x}^{2}}+Bx-B{{x}^{2}}+C-Cx$

We obtain the below values by equating the coefficients of x ${{x}^{2}}$ and the constant term on both sides. 

$\text{A}-\text{B}=0$

$B-C=0$

$A+C=2$

On solving these equations, to obtain

$A=1,B=1\text{, and }C=1$

$\therefore \dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{1}{1-x}+\dfrac{x+1}{1+{{x}^{2}}}$

$\Rightarrow \int{\dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}}dx=\int{\dfrac{1}{1-x}}dx+\int{\dfrac{x}{1+{{x}^{2}}}}dx+\int{\dfrac{1}{1+{{x}^{2}}}}dx$

$=-\int{\dfrac{1}{1-x}}dx+\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx+\int{\dfrac{1}{1+{{x}^{2}}}}dx$

$=-\log |x-1|+\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+{{\tan }^{-1}}x+C$


14. $\dfrac{3x-1}{{{\left( x+2 \right)}^{2}}}$

Ans:   Let $\dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{A}{(x+2)}+\dfrac{B}{{{(x+2)}^{2}}}$

$\Rightarrow 3x-1=A(x+2)+B$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$A=3$

$2A+B=-1\Rightarrow B=-7$

$\therefore \dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{3}{(x+2)}-\dfrac{7}{{{(x+2)}^{2}}}$

$\Rightarrow \int{\dfrac{3x-1}{{{(x+2)}^{2}}}}dx=3\int{\dfrac{1}{(x+2)}}dx-7\int{\dfrac{x}{{{(x+2)}^{2}}}}dx$

Using the power rule and logarithm formula of integration

$=3\log |x+2|-7\left( \dfrac{-1}{(x+2)} \right)+C$

$=3\log |x+2|+\dfrac{7}{(x+2)}+C$


15. $\dfrac{1}{{{x}^{4}}-1}$

Ans:  $\dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{1}{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}$

Let $\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{Cx+D}{\left( {{x}^{2}}+1 \right)}$ 

$1=A(x-1)\left( 1+{{x}^{2}} \right)+B(x+1)\left( 1+{{x}^{2}} \right)+(Cx+D)\left( {{x}^{2}}-1 \right)$

$1=A\left( {{x}^{3}}+x-{{x}^{2}}-1 \right)+B\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)+C{{x}^{3}}+D{{x}^{2}}-Cx-D$

$1=(A+B+C){{x}^{3}}+(-A+B+D){{x}^{2}}+(A+B-C)x+(-A+B-D)$

We obtain the below values by equating the coefficients of${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x},$and constant term, we get

$A+B+C=0\text{ }$

$-A+B+D=0$

$A+B-C=0\text{ }$

$-A+B-D=1$

$A=-\dfrac{1}{4},B=\dfrac{1}{4},C=0,\text{ and }D=-\dfrac{1}{2}$

$\therefore \dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{-1}{4(x+1)}+\dfrac{1}{4(x-1)}+\dfrac{1}{2\left( {{x}^{2}}+1 \right)}$

$\Rightarrow \int{\dfrac{1}{{{x}^{4}}-1}}dx=-\dfrac{1}{4}\log |x-1|+\dfrac{1}{4}\log |x-1|-\dfrac{1}{2}{{\tan }^{1}}x+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{x-1}{x+1} \right|-\dfrac{1}{2}{{\tan }^{1}}x+C$


16. $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$ $\text{ [ }$Hint: Multiply numerator and denominator by ${{x}^{n-1}}$ and put ${{x}^{n}}=t$ $\text{]}$

Ans: $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$

Numerator and denominator are multiplied by ${{x}^{n-1}}$, to obtain 

$\dfrac{1}{x\left( {{x}^{n}}+1 \right)}=\dfrac{{{x}^{n-1}}}{{{x}^{n-1}}x\left( {{x}^{n}}+1 \right)}=\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}+1 \right)}$

Consider  ${{x}^{n}}=t\Rightarrow {{x}^{n-1}}dx=dt$

$\therefore \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\int{\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}}dt$

$\text{ Let }\dfrac{1}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{(t+1)}$

$1=A\left( 1+t \right)+B\text{ }t$

We obtain the below values by equating the coefficients of $\text{t}$and constant, 

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{t(t+1)}=\dfrac{1}{t}-\dfrac{1}{(1+t)}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\left\{ \dfrac{1}{t}-\dfrac{1}{(1+t)} \right\}}dx$

$=\dfrac{1}{n}[\log |t|-\log |t+1|]+C$

Substitute the value of t,

$=-\dfrac{1}{n}\left[ \log \left| {{x}^{n}} \right|-\log \left| {{x}^{n}}+1 \right| \right]+C$

Simplifying,

$=\dfrac{1}{n}\log \left| \dfrac{{{x}^{u}}}{{{x}^{\prime \prime }}+1} \right|+C$


17. $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$ $\text{[}$ hint: Put $\sin x=t$ $\text{]}$

Ans:  $\dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,\,\,\,Put,\sin x=t\Rightarrow \cos xdx=dt$

$\therefore \int{\dfrac{\cos x}{(1-\sin x)(2-\sin x)}}dx=\int{\dfrac{dt}{(1-t)(2-t)}}$

let $\dfrac{1}{(1-t)(2-t)}=\dfrac{A}{(1-t)}+\dfrac{B}{(2-t)}$ 

$1=A\left( 2-t \right)+B\left( 1-t \right)$

We obtain the below values by equating the coefficients of t and constant, 

$-2~\text{A}-\text{B}=0\,\,,and\,\,2~\text{A}+\text{B}=1$

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{(1-t)(2-t)}=\dfrac{1}{(1-t)}-\dfrac{1}{(2-t)}$

$\Rightarrow \int{\dfrac{\cos x}{(1-\sin x)(2-\sin x)}}dx=\int{\left\{ \dfrac{1}{1-t}-\dfrac{1}{(2-t)} \right\}}dt=-\log |1-t|+\log |2-t|+C$ Simplifying,

$=\log \left| \dfrac{2-t}{1-t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{2-\sin x}{1-\sin x} \right|+C$


18. $\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}$

Ans:  $\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}$

Let $\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{Ax+B}{\left( {{x}^{2}}+3 \right)}+\dfrac{Cx+D}{\left( {{x}^{2}}+4 \right)}$

$4{{x}^{2}}+10=(Ax+B)\left( {{x}^{2}}+4 \right)+(Cx+D)\left( {{x}^{2}}+3 \right)$

$4{{x}^{2}}+10=A{{x}^{2}}+4Ax+B{{x}^{2}}+4B+C{{x}^{3}}+3Cx+D{{x}^{2}}+3D$

$4{{x}^{2}}+10=(A+C){{x}^{3}}+(B+D){{x}^{2}}+(4A+3C)x+(4B+3D)$

We obtain the below values by equating the coefficients of ${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x}$and constant term, 

$A+C=0$

$B+D=4$

$4\text{ }A+3\text{ }C=0$

$4\text{ }B+3\text{ }D=10$

On solving these equations, to obtain $\text{A}=0.\text{B}=-2.\text{C}=0,and\,\,\,\text{D}=6$

$\therefore \dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)}$

$\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\left( \dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)} \right)$

$\Rightarrow \int{\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}}dx=\int{\left\{ 1+\dfrac{2}{\left( {{x}^{2}}+3 \right)}-\dfrac{6}{\left( {{x}^{2}}+4 \right)} \right\}}dx$

$=\int{\left\{ 1+\dfrac{2}{{{x}^{2}}+{{(\sqrt{3})}^{2}}}-\dfrac{6}{{{x}^{2}}+{{2}^{2}}} \right\}}$

$=x+2\left( \dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\dfrac{x}{\sqrt{3}} \right)-6\left( \dfrac{1}{2}{{\tan }^{-1}}\dfrac{x}{2} \right)+C$

Simplifying,

$=x+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\dfrac{x}{\sqrt{3}}-3{{\tan }^{-1}}\dfrac{x}{2}+C$


19. $\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}$

Ans:  $\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}$

Put ${{x}^{2}}-t\to 2xdx-dt$

$\therefore \int{\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}}dx=\int{\dfrac{dt}{(t+1)(t+3)}}$

$\text{ Let }\dfrac{1}{(t+1)(t+3)}=\dfrac{A}{(t+1)}+\dfrac{B}{(t+3)}$

$I=A\left( t+3 \right)+B\left( t+1 \right)$

We obtain the below values by equating the coefficients of $\text{t}$and  

 constant, 

                                 $1+B=0$and $3A+B=1$

On solving, we get

$A=\dfrac{1}{2}\text{ and }B=-\dfrac{1}{2}$

$\therefore \dfrac{1}{(t+1)(t+3)}=\dfrac{1}{2(t+1)}+\dfrac{1}{2(t+3)}$

$\Rightarrow \int{\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}}dx=\int{\left\{ \dfrac{1}{2(t+1)}-\dfrac{1}{2(t+3)} \right\}}dt$

$=\dfrac{1}{2}\log |(t+1)|-\dfrac{1}{2}\log |t+3|+C$

Simplifying,

$=\dfrac{1}{2}\log \left| \dfrac{t+1}{t+3} \right|+C=\dfrac{1}{2}\log \left| \dfrac{{{x}^{2}}+1}{{{x}^{2}}+3} \right|+C$


20. $\dfrac{1}{x\left( {{x}^{4}}-1 \right)}$

 Ans:  $\dfrac{1}{x\left( {{x}^{4}}-1 \right)}$

Numerator and denominator are multiplied by by ${{\text{x}}^{3}},$we get

$\dfrac{1}{x\left( {{x}^{4}}-1 \right)}=\dfrac{{{x}^{3}}}{{{x}^{4}}\left( {{x}^{4}}-1 \right)}$

$\therefore \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\int{\dfrac{{{x}^{3}}}{{{x}^{4}}\left( {{x}^{4}}-1 \right)}}dx$

Consider ${{\text{x}}^{4}}=\text{t}\Rightarrow 4{{\text{x}}^{3}}\text{dx}=\text{dt}$

$\therefore \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\dfrac{1}{4}\int{\dfrac{dt}{t(t-1)}}$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{(t-1)}$

$1=A(t-1)+Bt\quad \ldots (1)$

We obtain the below values by equating the coefficients of $\text{t}$and constant, 

$A=-1\text{ and }B=1$

$\Rightarrow \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\dfrac{1}{4}\int{\left\{ \dfrac{-1}{t}+\dfrac{1}{t-1} \right\}}dt$

Using the logarithm formula of integration,

$=\dfrac{1}{4}[-\log |t|+\log |t-1|]+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{t-1}{t} \right|+C=\dfrac{1}{4}\log \left| \dfrac{{{x}^{4}}-1}{{{x}^{4}}} \right|+C$


21. $\dfrac{1}{{{e}^{x}}-1}$ $\text{[}$ hint: Put ${{e}^{x}}=t$ $\text{]}$

Ans:  $\dfrac{1}{\left( {{e}^{x}}-1 \right)}$

Put ${{\text{e}}^{x}}=\text{t }\Rightarrow {{\text{e}}^{x}}\text{dx}=dt$

$\Rightarrow \int{\dfrac{1}{\left( {{e}^{x}}-1 \right)}}dx=\int{\dfrac{1}{t-1}}\times \dfrac{dt}{t}=\int{\dfrac{1}{t(t-1)}}dt$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{t-1}$

$1=A\left( t-1 \right)+B\text{ }t$

We obtain the below values by equating the coefficients of t and constant, $A=-1\text{ and }B=1$

$\therefore \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int{\dfrac{1}{t(t-1)}}dt=\log \left| \dfrac{t-1}{t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{{{e}^{x}}-1}{{{e}^{x}}} \right|+C$


22. $\int{\dfrac{xdx}{\left( x-1 \right)\left( x-2 \right)}}$ equals

  1. $\log \left| \dfrac{{{\left( x-1 \right)}^{2}}}{x-2} \right|+C$

  2. $\log \left| \dfrac{{{\left( x-2 \right)}^{2}}}{x-1} \right|+C$

  3. $\log \left| {{\left( \dfrac{x-1}{x-2} \right)}^{2}} \right|+C$

  4. $\log \left| \left( x-1 \right)\left( x-2 \right) \right|+C$

 Ans:  Let $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}$

$x=A\left( x-2 \right)+B\left( x-1 \right)$

We obtain the below values by equating the coefficients of $\text{x}$and constant,  $\text{A}=-1\,\,\,and\,\,\text{B}=2$

$\therefore \dfrac{x}{(x-1)(x-2)}=-\dfrac{1}{(x1)}+\dfrac{2}{(x-2)}$

$\Rightarrow \int{\dfrac{x}{(x-1)(x-2)}}dx=\int{\left\{ \dfrac{-1}{(x-1)}+\dfrac{2}{(x-2)} \right\}}dx$

Using the logarithm formula of integration,

$=-\log |x-1|+2\log |x-2|+C$

Simplifying,

$=\log \left| \dfrac{{{(x-2)}^{2}}}{x-1} \right|+C$

Thus, the right response is B.


23. $\int{\dfrac{dx}{x\left( {{x}^{2}}+1 \right)}}$ equals 

  1. $\log \left| x \right|-\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

  2. $\log \left| x \right|+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

  3. $-\log \left| x \right|+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

  4. $\dfrac{1}{2}\log \left| x \right|+\log \left( {{x}^{2}}+1 \right)+C$

Ans: Let $\dfrac{1}{x\left( {{x}^{2}}+1 \right)}-\dfrac{A}{x},\dfrac{Bx+C}{{{x}^{2}}+1}$

$1=A\left( {{x}^{2}}+1 \right)+(Bx+C)x$

We obtain the below values by equating the coefficients of ${{\text{x}}^{2}},\text{x},$     and constant term, 

$A+B=0$, $C=0$$A=1$

On solving these equations, to obtain

$A=1,B=-1,\text{ and }C=0$

$\therefore \dfrac{1}{x\left( {{x}^{2}},1 \right)}=\dfrac{1}{x}+\dfrac{-x}{{{x}^{2}}+1}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{2}}+1 \right)}}dx=\int{\left\{ \dfrac{1}{x}-\dfrac{x}{{{x}^{2}}+1} \right\}}dx$

$=\log |x|-\dfrac{1}{2}\log \left| {{x}^{2}}+1 \right|+C$

Thus, the right response is $\text{A}.$


Exercise 7.6

1. $x\sin x$

 Ans:  Let $I=\int{x}\sin xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin \text{x}$and integrating by parts, to obtain  

$I=\int{x}\sin xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }xdx \right\}}dx$

$=x(-\cos x)-\int{1}.(-\cos x)dx$

$=-x\cos x+\sin x+C$


2. $x\sin 3x$

Ans: Let $\text{I}=\int{x}\sin 3xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin 3\text{x}$ and integrating by parts, to obtain  

$I=x\int{\sin }3xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }3xdx \right\}}$

$=x\left( \dfrac{-\cos 3x}{3} \right)-\int{1}\cdot \left( \dfrac{-\cos 3x}{3} \right)dx$

$=\dfrac{-x\cos 3x}{3}+\dfrac{1}{3}\int{\cos }3xdx=\dfrac{-x\cos 3x}{3}+\dfrac{1}{9}\sin 3x+C$


3. ${{x}^{2}}{{e}^{x}}$

Ans:  Let $I=\int{{{x}^{2}}}{{e}^{x}}dx$

Consider  $\text{u}={{\text{x}}^{2}}\,\,and\,\,\text{v}={{\text{e}}^{x}}$

$I={{x}^{2}}\int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx$

$={{x}^{2}}{{e}^{x}}-\int{2}x-{{e}^{x}}dx$

$={{x}^{2}}{{e}^{x}}-2\int{x}\cdot {{e}^{x}}dx$

Again using integration by parts, to obtain

$={{x}^{2}}{{e}^{x}}-2\left[ x\cdot \int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx \right]$

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-\int{{{e}^{x}}}dx \right]$

Simplifying,

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-{{e}^{x}} \right]$

$={{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+2{{e}^{x}}+C$

$={{e}^{x}}\left( {{x}^{2}}-2x+2 \right)+C$


4. $x\log x$

Ans:  Let $I=\int{x}\log xdx$

Consider $\text{u}=\log \text{x}\,\,\,\,and\,\,\,\text{v}=\text{x}$ and integrating by parts, to obtain

$I=\log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log x}{2}\cdot \sqrt{\dfrac{x}{2}}dx=\dfrac{{{x}^{2}}\log x}{2}-\dfrac{{{x}^{2}}}{4}+C$


5. $x\log 2x$

 Ans:  Let $I=\int{x}\log 2xdx$

Consider $u=\log 2x$ and $v=x$and integrating by parts, to obtain

$I=\log 2x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}2\log x \right)\int{x}dx \right\}}dx$

$=\log 2x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{2}{2x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log 2x}{2}-\int{\dfrac{x}{2}}dx$

Integrating using the power rule

$=\dfrac{{{x}^{2}}\log 2x}{2}-\dfrac{{{x}^{2}}}{4}+C$


6. ${{x}^{2}}\log x$

 Ans: Let $I=\int{{{x}^{2}}}\log xdx$

Consider $u=\log x$and $v={{x}^{2}}$ and integrating by parts, to obtain

$I=\log x\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

$=\log x\left( \dfrac{{{x}^{3}}}{3} \right)-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx$

Integrating using the power rule

$=\dfrac{{{x}^{3}}\log x}{3}-\int{\dfrac{{{x}^{2}}}{3}}dx=\dfrac{{{x}^{3}}\log x}{3}-\dfrac{{{x}^{3}}}{9}+C$


7. $x{{\sin }^{-1}}x$

 Ans: Let $I=\int{x}{{\sin }^{-1}}xdx$

Consider $u={{\sin }^{-1}}x\,\,and\,\,\,v=x$ and integrating by parts, to obtain

$I={{\sin }^{-1}}x\int{x}dx\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

Adding and subtracting by 1

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

Simplifying,

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \int{\sqrt{1-{{x}^{2}}}}dx-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right\}$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x-{{\sin }^{-1}}x \right\}+C$

Simplifying,      $=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}x-\dfrac{1}{2}{{\sin }^{-1}}x+C=\dfrac{1}{4}\left( 2{{x}^{2}}-1 \right){{\operatorname{in}}^{-1}}x+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C$


8. $x{{\tan }^{-1}}x$

Ans: Let $I=\int{x}{{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)\int{\dfrac{1}{1+{{x}^{2}}}}\cdot \dfrac{{{x}^{2}}}{2}dx=\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{1+{{x}^{2}}}}dx$

Adding and subtracting by -1

$=\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\left( \dfrac{{{x}^{2}}+1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}} \right)}dx=\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\left( 1-\dfrac{1}{1+{{x}^{2}}} \right)}dx$

Simplifying,

$=\dfrac{{{x}^{2}}{{\operatorname{lan}}^{-1}}x}{2}-\dfrac{1}{2}\left( x-{{\tan }^{-1}}x \right)+C=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{x}{2}+\dfrac{1}{2}{{\tan }^{-1}}x+C$


9. $x{{\cos }^{-1}}x$

 Ans:Let $I=\int{x}{{\cos }^{-1}}xdx$

 Taking $u={{\cos }^{-1}}x$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\cos }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\cos }^{-1}}x\dfrac{{{x}^{2}}}{2}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx$

Adding and subtracting by -1

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx$

Simplifying,

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}+\left( \dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right) \right\}}dx$

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\sqrt{1-{{x}^{2}}}}dx-\dfrac{1}{2}\int{\left( \dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right)}dx$

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}{{I}_{1}}-\dfrac{1}{2}{{\cos }^{-1}}x....\left( 1 \right)$

Where ${{I}_{1}}=\int{\sqrt{1-{{x}^{2}}}}dx$

$\Rightarrow {{I}_{1}}=x\int{\sqrt{1-{{x}^{2}}}}-\int{\dfrac{d}{dx}}\sqrt{1-{{x}^{2}}}\int{x}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ \int{\sqrt{1-{{x}^{2}}}}dx+\int{\dfrac{-dx}{\sqrt{1-{{x}^{2}}}}} \right\}$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ {{I}_{1}}+{{\cos }^{-1}}x \right\}\Rightarrow 2{{I}_{1}}=x\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x$

$\therefore {{I}_{1}}=\dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x$

Substituting in (1), we get

$I=\dfrac{x{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\left( \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x \right)-\dfrac{1}{2}{{\cos }^{-1}}x$

Simplifying,

$=\dfrac{\left( 2{{x}^{2}}-1 \right)}{4}{{\cos }^{-1}}x-\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C$


10. ${{\left( {{\sin }^{-1}}x \right)}^{2}}$

Ans: Let $I=\int{{{\left( {{\sin }^{-1}}x \right)}^{2}}}\cdot 1dx$

Consider $\text{u}={{\left( {{\sin }^{-1}}\text{x} \right)}^{2}}$ and $\text{v}=1$and integrating by parts, to obtain

$I=\int{\left( {{\sin }^{-1}}x \right)}\cdot \int{1}dx-\int{\left\{ \dfrac{d}{dx}{{\left( {{\sin }^{-1}}x \right)}^{2}}\cdot \int{1}.dx \right\}}dx$

$={{\left( {{\sin }^{-1}}x \right)}^{2}}x-\int{\dfrac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}\cdot xdx$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\int{{{\sin }^{-1}}}x\cdot \left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)dx$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-\int{2}dx$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-2x+C$


11. $\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$

Ans:  Let $I=\int{\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

Multiplying and dividing by 2

$I=\dfrac{-1}{2}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\cdot {{\cos }^{-1}}xdx$

Consider  $\text{u}={{\cos }^{-1}}\text{x}$ and $\text{v}=\left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)$and integrating by parts, to obtain

$I=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+\int{2}dx \right]$

Simplifying,

$=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+2x \right]+C$

$=-\left[ \sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+x \right]+C$


12. $x{{\sec }^{2}}x$

 Ans:Let $I=\int{x}{{\sec }^{2}}xdx$

Consider$\text{u}=\text{x}$ and $\text{v}={{\sec }^{2}}\text{x}$ and integrating by parts, to obtain

$I=x\int{{{\sec }^{2}}}xdx-\int{\left\{ \left\{ \dfrac{d}{dx}x \right\}\int{{{\sec }^{2}}}xdx \right\}}dx$

$=x\tan x-\int{1}\cdot \tan xdx$

$=x\tan x+\log |\cos x|+C$


13. ${{\tan }^{-1}}x$

Ans: Let $I=\int{1}\cdot {{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=1$ and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{1}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{1}.dx \right\}}dx={{\tan }^{-1}}xx-\int{\dfrac{1}{1+{{x}^{2}}}}xd$

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx$

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+C$ 

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C$


14. $x{{\left( \log x \right)}^{2}}dx$

Ans:  $I=\int{x}{{(\log x)}^{2}}dx$

Consider$u={{(\log x)}^{2}}$ and $v=1$ and integrating by parts, to obtain

$I={{(\log )}^{2}}\int{x}dx-\int{\left[ \left\{ {{\left( \dfrac{d}{dx}\log x \right)}^{2}} \right\}\int{x}dx \right]}dx$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \int{2}\log x\cdot \dfrac{1}{x}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

Again using integration by parts, to obtain

$I=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \dfrac{{{x}^{2}}}{2}-\log x-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{1}{2}\int{x}dx=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{{{x}^{2}}}{4}+C$


15. $\left( {{x}^{2}}+1 \right)\log x$

 Ans: Let $I=\int{\left( {{x}^{2}}+1 \right)}\log xdx=\int{{{x}^{2}}}\log xdx+\int{\log }xdx$

Let $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}\ldots (1)$

Where,${{I}_{1}}=\int{{{x}^{2}}}\log xdx\,\,\,\,\,\,and\,\,\,{{\text{I}}_{2}}=\int{\log }xdx$

${{I}_{1}}=\int{{{x}^{2}}}\log xdx$ 

Consider$\text{u}=\log \text{x}$ and $v=x^2$ and integrating by parts, to obtain

${{I}_{1}}=\log x-\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{3}}}{3}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{1}{3}\int{{{x}^{2}}}dx$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}\quad \ldots (2)$

${{I}_{2}}=\int{\log }xdx$

Consider $\text{u}=\log \text{x}$ and $\text{v}=1$and integrating by parts, to obtain

${{I}_{2}}=\log x\int{1}.dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{1}.dx \right\}}$

$=\log x\cdot x-\int{\dfrac{1}{x}}xdx$

$=x\log x-x..\left( 3 \right)$

Using equations (2) and (3) in (1), we get

$I=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}+x\log x-x+{{C}_{2}}$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+x\log x-x+\left( {{C}_{1}}+{{C}_{2}} \right)$

$=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C$


16. ${{e}^{x}}\left( \sin x+\cos x \right)$

Ans: Consider$I=\int{{{e}^{x}}}(\sin x+\cos x)dx$

Consider$f(x)=\sin x$

${{f}^{\prime }}(x)=\cos x$

$I=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore I={{e}^{x}}\sin x+C$


17. $\dfrac{x{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}$

Ans:  Consider $I=\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ \dfrac{x}{{{(1+x)}^{2}}} \right\}dx$

$=\int{{{e}^{x}}}\left\{ \dfrac{1+x-1}{{{(1+x)}^{2}}} \right\}dx=\int{{{e}^{x}}}\left\{ \dfrac{1}{1+x}-\dfrac{1}{{{(1+x)}^{2}}} \right\}dx$

Here, $f(x)=\dfrac{1}{1+x}\quad {{f}^{\prime }}(x)=\dfrac{-1}{{{(1+x)}^{2}}}$

$\Rightarrow \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\dfrac{{{e}^{x}}}{1+x}+C$


18. Integrate the function - ${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

Ans: First simplify –${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

It is known that – 

$1+\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

$1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$

$\therefore {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$={{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \tan \dfrac{x}{2}+1 \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\tan }^{2}}\dfrac{x}{2}+1+2\tan \dfrac{x}{2} \right)$

But, $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right)$

$={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

$\Rightarrow {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

If we say, $f(x)=\tan \dfrac{x}{2}\Rightarrow f'(x)=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$

Thus, we get – $\int{{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)}dx={{e}^{x}}\tan \dfrac{x}{2}+C$


19. Integrate the function - ${{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)$

Ans: Say, $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}$

Suppose, $f(x)=\dfrac{1}{x}\Rightarrow f'(x)=-\dfrac{1}{{{x}^{2}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, we get – $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}=\dfrac{{{e}^{x}}}{x}+C$


20. Integrate the function - $\dfrac{(x-3){{e}^{x}}}{{{(x-1)}^{3}}}$

Ans: $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\int{{{e}^{x}}\left[ \dfrac{(x-1-2)}{{{(x-1)}^{3}}} \right]dx}}$

$=\int{{{e}^{x}}\left[ \dfrac{(x-1)}{{{(x-1)}^{3}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx}$

$=\int{{{e}^{x}}\left[ \dfrac{1}{{{(x-1)}^{2}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx}$

Suppose, $f(x)=\dfrac{1}{{{(x-1)}^{2}}}\Rightarrow f'(x)=-\dfrac{2}{{{(x-1)}^{3}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\dfrac{{{e}^{x}}}{{{(x-1)}^{2}}}+C}$


21. Integrate the function - ${{e}^{2x}}\sin x$

Ans: Say,  $I=\int{{{e}^{2x}}\sin xdx}$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\sin x\text{   }v={{e}^{2x}}$

$I=\int{{{e}^{2x}}\sin x}dx=\sin x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\sin x \right)\int{{{e}^{2x}}dx} \right]dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( \cos x \right)\dfrac{{{e}^{2x}}}{2} \right]dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\int{\left( {{e}^{2x}}\cos x \right)dx}$

Perform Integration by parts for – $\int{\left( {{e}^{2x}}\cos x \right)dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\cos x \right)\int{{{e}^{2x}}dx} \right]dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( -\sin x \right)\dfrac{{{e}^{2x}}}{2} \right]dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}+\dfrac{1}{2}\int{(\sin x){{e}^{2x}}dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}\left\{ \int{(\sin x){{e}^{2x}}dx} \right\}$

But, $I=\int{{{e}^{2x}}\sin xdx}$

$\Rightarrow I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}I$

$\Rightarrow I+\dfrac{1}{4}I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{2{{e}^{2x}}\sin x}{4}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow 5I={{e}^{2x}}(2\sin x-\cos x)$

Thus, we get – $I=\dfrac{{{e}^{2x}}}{5}(2\sin x-\cos x)+C$.


22. Integrate the function - ${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)$

Ans: Say, $x=\tan \theta \text{ }\Rightarrow \text{dx=se}{{\text{c}}^{2}}\theta d\theta $

$\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)$

But, $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta }$

$\Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)=\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \sin 2\theta  \right)=2\theta $

Therefore, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=\int{2\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

$=2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\theta \text{   }v={{\sec }^{2}}\theta $

$2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }=2\left\{ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left[ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -\int{\left[ \tan \theta  \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -(-\log |\cos \theta |) \right\}+C$

$=2\left\{ \theta \tan \theta +\log |\cos \theta | \right\}+C$

Replace $\theta ={{\tan }^{-1}}x$

$=2\left\{ {{\tan }^{-1}}x\tan ({{\tan }^{-1}}x)+\log |\cos ({{\tan }^{-1}}x)| \right\}+C$

It is known that – ${{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}$

$=2\left\{ {{\tan }^{-1}}x(x)+\log |\cos ({{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}})| \right\}+C$

$=2\left\{ x{{\tan }^{-1}}x+\log |\dfrac{1}{\sqrt{1+{{x}^{2}}}}| \right\}+C$

$=2\left\{ x{{\tan }^{-1}}x+\log {{(1+{{x}^{2}})}^{-\dfrac{1}{2}}} \right\}+C$

Here, $\log {{m}^{n}}=n\log m$

$=2\left\{ x{{\tan }^{-1}}x-\dfrac{1}{2}\log (1+{{x}^{2}}) \right\}+C$

$=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

Thus, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$


23. Choose the correct answer: $\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$ equals

  1. $\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

  2. $\dfrac{1}{3}{{e}^{{{x}^{2}}}}+C$

  3. $\dfrac{1}{2}{{e}^{{{x}^{3}}}}+C$

  4. $\dfrac{1}{2}{{e}^{{{x}^{2}}}}+C$

Ans: Say, $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

Suppose, $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx$

Rewriting the equation – $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx=\dfrac{1}{3}\int{{{e}^{t}}}dt$

$\Rightarrow I=\dfrac{1}{3}\int{{{e}^{t}}}dt=\dfrac{1}{3}{{e}^{t}}+C$

Replacing $t={{x}^{3}}$

$\Rightarrow I=\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

The correct option is A.


24. Choose the correct answer: $\int{{{e}^{x}}\sec x(1+\tan x)}dx$

  1. ${{e}^{x}}\cos x+C$

  2. ${{e}^{x}}\sec x+C$

  3. ${{e}^{x}}\sin x+C$

  4. ${{e}^{x}}\tan x+C$

Ans: Say, $I=\int{{{e}^{x}}\sec x(1+\tan x)}dx$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx$

Suppose, $f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx={{e}^{x}}\sec x+C$

Thus, $I={{e}^{x}}\sec x+C$

The correct option is B.


Exercise- 7.7

1. Integrate the function - $\sqrt{4-{{x}^{2}}}$

Ans: Say, $I=\int{\sqrt{4-{{x}^{2}}}dx}=\int{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow I=\int{\sqrt{{{2}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{2}^{2}}-{{x}^{2}}}+\dfrac{{{2}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{2}+C}$

$\Rightarrow I=\dfrac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\dfrac{x}{2}+C$

Thus, $\int{\sqrt{4-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\dfrac{x}{2}+C$


2. Integrate the function - $\sqrt{1-4{{x}^{2}}}$

Ans: Say, $I=\int{\sqrt{1-4{{x}^{2}}}dx}=\int{\sqrt{{{1}^{2}}-{{(2x)}^{2}}}dx}$

Let, $2x=t\Rightarrow 2dx=dt$

$x=\dfrac{t}{2}\Rightarrow dx=\dfrac{dt}{2}$

So, we get – $I=\int{\sqrt{{{1}^{2}}-{{\left[ 2(\dfrac{t}{2}) \right]}^{2}}}\dfrac{dt}{2}}=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt}$

$\Rightarrow I=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{t}{2}\sqrt{1-{{t}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}t \right]+C$

$\Rightarrow I=\left[ \dfrac{t}{4}\sqrt{1-{{t}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}t \right]+C$

Replace – $t=2x$

$\Rightarrow I=\left[ \dfrac{2x}{4}\sqrt{1-{{(2x)}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C$

$\Rightarrow I=\left[ \dfrac{x}{2}\sqrt{1-4{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C$

Thus,$\int{\sqrt{1-4{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{1-4{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x+C$


3. Integrate the function - $\sqrt{{{x}^{2}}+4x+6}$

Ans: First simplify –${{x}^{2}}+4x+6$

${{x}^{2}}+4x+6={{x}^{2}}+4x+4+2$

$=({{x}^{2}}+4x+4)+2={{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x+6}=\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x+6}dx}=\int{\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}+\dfrac{{{(\sqrt{2})}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+6}+\dfrac{2}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+6} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+4x+6}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+6}+\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+6} \right|+C$


4. Integrate the function - $\sqrt{{{x}^{2}}+4x+1}$

Ans: First simplify –${{x}^{2}}+4x+1$

${{x}^{2}}+4x+1={{x}^{2}}+4x+4-3$

$=({{x}^{2}}+4x+4)-3={{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x+1}=\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x+1}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}-\dfrac{{{(\sqrt{3})}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+1}-\dfrac{3}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+1} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+4x+1}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+1}-\dfrac{3}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+1} \right|+C$


5. Integrate the function - $\sqrt{1-4x-{{x}^{2}}}$

Ans: First simplify –$1-4x-{{x}^{2}}$

$1-4x-{{x}^{2}}=1-4x-{{x}^{2}}-4+4=1+4-({{x}^{2}}+4x+4)$

$=5-({{x}^{2}}+4x+4)={{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}$

$\Rightarrow \sqrt{1-4x-{{x}^{2}}}=\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}$

$\therefore \int{\sqrt{1-4x-{{x}^{2}}}dx}=\int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow \int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}+\dfrac{{{(\sqrt{5})}^{2}}}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$$=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$

Thus, $\int{\sqrt{1-4x-{{x}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$


6. Integrate the function - $\sqrt{{{x}^{2}}+4x+5}$

Ans: First simplify –${{x}^{2}}+4x-5$

${{x}^{2}}+4x-5={{x}^{2}}+4x-5+4-4=({{x}^{2}}+4x+4)-5-4$

$=({{x}^{2}}+4x+4)-9={{(x+2)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x-5}=\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x-5}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}-\dfrac{{{(3)}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x-5}-\dfrac{9}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x-5} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+4x-5}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x-5}-\dfrac{9}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x-5} \right|+C$


7. Integrate the function - $\sqrt{1+3x-{{x}^{2}}}$

Ans: First simplify –$1+3x-{{x}^{2}}$

$1+3x-{{x}^{2}}=1-{{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=1+\dfrac{9}{4}-({{x}^{2}}-3x+\dfrac{9}{4})$

$=\dfrac{9+4}{4}-({{x}^{2}}-3x+\dfrac{9}{4})=\left( \dfrac{13}{4} \right)-({{x}^{2}}-3x+\dfrac{9}{4})={{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}$

$\Rightarrow \sqrt{1+3x-{{x}^{2}}}=\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}$

$\therefore \int{\sqrt{1+3x-{{x}^{2}}}dx}=\int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow \int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x-\dfrac{3}{2} \right)}{2}\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}+\dfrac{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{\left( x-\dfrac{3}{2} \right)}{\left( \dfrac{\sqrt{13}}{2} \right)}+C$$=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$

Thus, $\int{\sqrt{1+3x-{{x}^{2}}}dx}=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$


8. Integrate the function - $\sqrt{{{x}^{2}}+3x}$

Ans: First simplify –${{x}^{2}}+3x$

${{x}^{2}}+x={{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=({{x}^{2}}+3x+\dfrac{9}{4})-\dfrac{9}{4}$

$={{\left( x+\dfrac{3}{2} \right)}^{2}}-\left( \dfrac{9}{4} \right)={{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+3x}=\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+3x}dx}=\int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x+\dfrac{3}{2} \right)}{2}\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}-\dfrac{{{\left( \dfrac{3}{2} \right)}^{2}}}{2}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}} \right|+C$$=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+3x}dx}=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C$


9. Integrate the function - $\sqrt{1+\dfrac{{{x}^{2}}}{9}}$

Ans: First simplify –$1+\dfrac{{{x}^{2}}}{9}$

$1+\dfrac{{{x}^{2}}}{9}=\dfrac{1}{9}(9+{{x}^{2}})=\dfrac{1}{9}({{3}^{2}}+{{x}^{2}})$

$\Rightarrow \sqrt{1+\dfrac{{{x}^{2}}}{9}}=\sqrt{\dfrac{1}{9}({{3}^{2}}+{{x}^{2}})}=\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})}$

$\therefore \int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\int{\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})}dx}=\dfrac{1}{3}\int{\sqrt{({{3}^{2}}+{{x}^{2}})}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

$\Rightarrow \dfrac{1}{3}\int{\sqrt{({{3}^{2}}+{{x}^{2}})}dx}=\dfrac{1}{3}\left\{ \dfrac{x}{2}\sqrt{{{(x)}^{2}}+{{(3)}^{2}}}+\dfrac{{{(3)}^{2}}}{2}\log \left| (x)+\sqrt{{{(x)}^{2}}+{{(3)}^{2}}} \right| \right\}+C$$=\dfrac{1}{3}\left\{ \dfrac{x}{2}\sqrt{{{x}^{2}}+9}+\dfrac{9}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right| \right\}+C$

$\dfrac{x}{6}{\sqrt{x^2+9}}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right| +C$
Thus, $\int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\dfrac{x}{6}\sqrt{{{x}^{2}}+9}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right|+C$


10. Choose the correct answer: $\int{\sqrt{1+{{x}^{2}}}dx}$ is equal to –

  1. $\dfrac{x}{2}\sqrt{1+{{x}^{2}}}+\dfrac{1}{2}\log \left| (x+\sqrt{1+{{x}^{2}}}) \right|+C$

  2. $\dfrac{2}{3}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

  3. $\dfrac{2}{3}x{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

  4. $\dfrac{{{x}^{2}}}{2}\sqrt{1+{{x}^{2}}}+\dfrac{1}{2}{{x}^{2}}\log \left| (x+\sqrt{1+{{x}^{2}}}) \right|+C$

Ans: It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

Thus,  $\int{\sqrt{{{x}^{2}}+{{1}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{1}^{2}}}+\dfrac{{{1}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{1}^{2}}} \right|+C}$

$\int{\sqrt{{{x}^{2}}+1}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+1}+\dfrac{1}{2}\log \left| x+\sqrt{{{x}^{2}}+1} \right|+C}$

The correct answer is option A.


11. Choose the correct answer: $\int{\sqrt{{{x}^{2}}-8x+7}dx}$ is equal to –

  1. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

  2. $\dfrac{1}{2}(x+4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x+4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

  3. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}-3\sqrt{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

  4. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+\dfrac{9}{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

Ans: First simplify –${{x}^{2}}-8x+7$

${{x}^{2}}-8x+7+9-9={{x}^{2}}-8x+16-9=({{x}^{2}}-8x+16)-9$

$={{(x-4)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}-8x+7}=\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}-8x+7}dx}=\int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}=\dfrac{(x-4)}{2}\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}-\dfrac{{{(3)}^{2}}}{2}\log \left| (x-4)+\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}} \right|+C$$=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}-8x+7}dx}=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

The correct answer is option D


Exercise- 7.8

1. Evaluate the definite integral as limit of sum – $\int\limits_{a}^{b}{xdx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x\text{ ; }a=a\text{ ; }b=b\text{ ; }h=\dfrac{b-a}{n}$

 $\therefore \int\limits_{a}^{b}{xdx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ a+(a+h)+(a+2h)+...+(a+(n-1)h) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(a+a+a+...+a)}_{n-times}}+(h+2h+...+(n-1)h) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ na+h(1+2+3+...+(n-1)) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ na+h\left( \dfrac{n(n-1)}{2} \right) \right]$

‘n’ is a common factor

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ a+\left( \dfrac{h(n-1)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{h(n-1)}{2} \right) \right]$

Replace the value for ‘h’

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{\left( \dfrac{b-a}{n} \right)(n-1)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)(n-1)}{2n} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=(b-a)\left[ a+\left( \dfrac{(b-a)\left( 1-0 \right)}{2} \right) \right]$

$=(b-a)\left[ a+\left( \dfrac{b-a}{2} \right) \right]$

$=(b-a)\left[ \left( \dfrac{2a+b-a}{2} \right) \right]$

$=(b-a)\left( \dfrac{a+b}{2} \right)$

$=\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{2} \right)$

Thus, $\int\limits_{a}^{b}{xdx}=\dfrac{1}{2}({{b}^{2}}-{{a}^{2}})$


2. Evaluate the definite integral as limit of sum – $\int\limits_{0}^{5}{(x+1)dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x+1\text{ ; }a=0\text{ ; }b=5\text{ ; }h=\dfrac{b-a}{n}=\dfrac{5-0}{n}=\dfrac{5}{n}$

 $\therefore \int\limits_{0}^{5}{(x+1)dx}=(5-0)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(0)+f\left( \dfrac{5}{n} \right)+f\left( \dfrac{10}{n} \right)+...+f\left( \dfrac{(n-1)5}{n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+\left( \dfrac{5}{n}+1 \right)+\left( \dfrac{10}{n}+1 \right)+...+\left( \dfrac{(n-1)5}{n}+1 \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(1+1+1...+1)}_{n-times}}+\left( \dfrac{5}{n} \right)+\left( \dfrac{(2)5}{n} \right)+...+\left( \dfrac{(n-1)5}{n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{5}{n}(1+2+3+...+(n-1) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{5}{n}\left( \dfrac{n(n-1)}{2} \right) \right]$

‘n’ is a common factor

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ 1+\left( \dfrac{5(n-1)}{2n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5(n-1)}{2n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=5\left[ 1+\left( \dfrac{5\left( 1-0 \right)}{2} \right) \right]$

$=5\left[ 1+\left( \dfrac{5}{2} \right) \right]$

$=5\left[ \dfrac{7}{2} \right]$

$=\left( \dfrac{35}{2} \right)$

Thus, $\int\limits_{0}^{5}{(x+1)dx}=\dfrac{35}{2}$


3. Evaluate the definite integral as limit of sum – $\int\limits_{2}^{3}{{{x}^{2}}dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)={{x}^{2}}\text{ ; }a=2\text{ ; }b=3\text{ ; }h=\dfrac{3-2}{n}=\dfrac{1}{n}$

 $\therefore \int\limits_{2}^{3}{{{x}^{2}}dx}=(3-2)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(2)+f\left( 2+\dfrac{1}{n} \right)+f\left( 2+\dfrac{2}{n} \right)+...+f\left( 2+\dfrac{(n-1)}{n} \right) \right]$

$=1\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}+{{\left( 2+\dfrac{1}{n} \right)}^{2}}+{{\left( 2+\dfrac{2}{n} \right)}^{2}}+...+{{\left( 2+\dfrac{(n-1)}{n} \right)}^{2}} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}+\left\{ {{2}^{2}}+\dfrac{1}{{{n}^{2}}}+2.2.\dfrac{1}{n} \right\}+\left\{ {{2}^{2}}+\dfrac{{{2}^{2}}}{{{n}^{2}}}+2.2.\dfrac{2}{n} \right\}+...+\left\{ {{2}^{2}}+\dfrac{{{(n-1)}^{2}}}{{{n}^{2}}}+2.2.\dfrac{(n-1)}{n} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{\left\{ {{2}^{2}}+{{2}^{2}}+{{...2}^{2}} \right\}}_{n-times}}+\left\{ \dfrac{1}{{{n}^{2}}}+\dfrac{{{2}^{2}}}{{{n}^{2}}}+...+\dfrac{{{(n-1)}^{2}}}{{{n}^{2}}} \right\}+\left\{ 2.2.\dfrac{1}{n}+2.2.\dfrac{2}{n}+...+2.2.\dfrac{(n-1)}{n} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}n+\dfrac{1}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+2.2.\dfrac{1}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+\dfrac{4}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

It is also known that the sum of n-squared-terms is – ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{(n-1)}^{2}}=\dfrac{n(n-1)(2n-1)}{6}$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{{{n}^{2}}}\left\{ \dfrac{n(n-1)(2n-1)}{6} \right\}+\dfrac{4}{n}\left\{ \dfrac{n(n-1)}{2} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{n}\left\{ \dfrac{(n-1)(2n-1)}{6} \right\}+4\left\{ \dfrac{(n-1)}{2} \right\} \right]$

Taking $\dfrac{1}{n}$ inside

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\dfrac{1}{{{n}^{2}}}\left\{ \dfrac{(n-1)(2n-1)}{6} \right\}+\dfrac{4}{n}\left\{ \dfrac{(n-1)}{2} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\left\{ \dfrac{\left( \dfrac{(n-1)}{n} \right)\left( \dfrac{(2n-1)}{n} \right)}{6} \right\}+2\left\{ \left( \dfrac{(n-1)}{n} \right) \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\left\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 2-\dfrac{1}{n} \right)}{6} \right\}+2\left\{ \left( 1-\dfrac{1}{n} \right) \right\} \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=\left[ 4+\left\{ \dfrac{\left( 1-0 \right)\left( 2-0 \right)}{6} \right\}+2\left\{ 1-0 \right\} \right]$

$=\left[ 4+\left\{ \dfrac{\left( 1 \right)\left( 2 \right)}{6} \right\}+2\left\{ 1 \right\} \right]$

$=\left[ 4+\left\{ \dfrac{1}{3} \right\}+2 \right]$

$=\dfrac{12+1+6}{3}$

$=\dfrac{19}{3}$

Thus, $\int\limits_{2}^{3}{{{x}^{2}}dx}=\dfrac{19}{3}$


4. Evaluate the definite integral as limit of sum – $\int\limits_{1}^{4}{({{x}^{2}}-x)dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Let us say $I={{I}_{1}}-{{I}_{2}}$

With, ${{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}\text{  ;  }{{I}_{2}}=\int\limits_{1}^{4}{xdx}$

Finding ${{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}$

Here, $f(x)={{x}^{2}}\text{ ; }a=1\text{ ; }b=4\text{ ; }h=\dfrac{4-1}{n}=\dfrac{3}{n}$

$\therefore {{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}=(4-1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(1)+f\left( 1+\dfrac{3}{n} \right)+f\left( 1+\dfrac{(2)3}{n} \right)+...+f\left( 1+\dfrac{(n-1)3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}+{{\left( 1+\dfrac{3}{n} \right)}^{2}}+{{\left( 1+\dfrac{6}{n} \right)}^{2}}+...+{{\left( 1+\dfrac{3(n-1)}{n} \right)}^{2}} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}+\left\{ {{1}^{2}}+\dfrac{{{3}^{2}}}{{{n}^{2}}}+2.2.\dfrac{3}{n} \right\}+\left\{ {{1}^{2}}+\dfrac{{{2}^{2}}{{.3}^{2}}}{{{n}^{2}}}+2.2.\dfrac{2.3}{n} \right\}+...+\left\{ {{1}^{2}}+\dfrac{{{3}^{2}}{{(n-1)}^{2}}}{{{n}^{2}}}+2.2.\dfrac{(n-1)3}{n} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{\left\{ {{1}^{2}}+{{1}^{2}}+{{...1}^{2}} \right\}}_{n-times}}+\left\{ \dfrac{{{3}^{2}}}{{{n}^{2}}}+\dfrac{{{2}^{2}}{{.3}^{2}}}{{{n}^{2}}}+...+\dfrac{{{(n-1)}^{2}}{{.3}^{2}}}{{{n}^{2}}} \right\}+\left\{ 2.\dfrac{3}{n}+2.\dfrac{2.3}{n}+...+2.\dfrac{(n-1).3}{n} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}n+\dfrac{{{3}^{2}}}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+2.\dfrac{3}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{9}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+\dfrac{6}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

It is also known that the sum of n-squared-terms is – ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{(n-1)}^{2}}=\dfrac{n(n-1)(2n-1)}{6}$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{9}{{{n}^{2}}}\left\{ \dfrac{n(n-1)(2n-1)}{6} \right\}+\dfrac{6}{n}\left\{ \dfrac{n(n-1)}{2} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}\left\{ \dfrac{(n-1)(2n-1)}{2} \right\}+3(n-1) \right]$

Taking $\dfrac{1}{n}$ inside

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\dfrac{3}{{{n}^{2}}}\left\{ \dfrac{(n-1)(2n-1)}{2} \right\}+\dfrac{3}{n}(n-1) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+3\left\{ \dfrac{\left( \dfrac{(n-1)}{n} \right)\left( \dfrac{(2n-1)}{n} \right)}{2} \right\}+3\left( \dfrac{(n-1)}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+3\left\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 2-\dfrac{1}{n} \right)}{2} \right\}+3\left( 1-\dfrac{1}{n} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=3\left[ 1+3\left\{ \dfrac{\left( 1-0 \right)\left( 2-0 \right)}{2} \right\}+3\left( 1-0 \right) \right]$

$=3\left[ 1+3\left\{ \dfrac{\left( 1 \right)\left( 2 \right)}{2} \right\}+3\left( 1 \right) \right]$

$=3\left[ 1+3+3 \right]$

$=3[7]$

$\therefore {{I}_{1}}=21$

Finding ${{I}_{2}}=\int\limits_{1}^{4}{xdx}$

Here, $f(x)=x\text{ ; }a=1\text{ ; }b=4\text{ ; }h=\dfrac{4-1}{n}=\dfrac{3}{n}$

 $\therefore {{I}_{2}}=\int\limits_{1}^{4}{xdx}=(4-1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(1)+f\left( 1+\dfrac{3}{n} \right)+f\left( 1+\dfrac{2.3}{n} \right)+...+f\left( 1+\dfrac{(n-1).3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+\left( 1+\dfrac{3}{n} \right)+\left( 1+\dfrac{2.3}{n} \right)+...+\left( 1+\dfrac{(n-1).3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(1+1+1+...+1)}_{n-times}}+\left( \dfrac{3}{n}+2.\dfrac{3}{n}+...+(n-1).\dfrac{3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}(1+2+3+...+(n-1)) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}\left( \dfrac{n\left( n-1 \right)}{2} \right) \right]$

‘n’ is a common factor

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ 1+\dfrac{3}{n}\left( \dfrac{(n-1)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\dfrac{3}{n}\left( \dfrac{(n-1)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{3\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{3\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=3\left[ 1+\left( \dfrac{3\left( 1-0 \right)}{2} \right) \right]$

$=3\left[ 1+\left( \dfrac{3}{2} \right) \right]$

$=3\left( \dfrac{2+3}{2} \right)$

$=3\left( \dfrac{5}{2} \right)$

$=\dfrac{15}{2}$

$\therefore {{I}_{2}}=\dfrac{15}{2}$

Now, $I={{I}_{1}}-{{I}_{2}}$

$\Rightarrow I=21-\dfrac{15}{2}=\dfrac{42-15}{2}=\dfrac{27}{2}$

Thus, $\int\limits_{1}^{4}{({{x}^{2}}-x)dx}=\dfrac{27}{2}$


5. Evaluate the definite integral as limit of sum – $\int\limits_{-1}^{1}{{{e}^{x}}dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,}\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+...+f\left( a+\left( n-1 \right)h \right) \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)={{e}^{x}}\text{ ; }a=-1\text{ ; }b=1\text{ ; }h=\dfrac{1-(-1)}{n}=\dfrac{1+1}{n}=\dfrac{2}{n}$

 $\therefore \int\limits_{-1}^{1}{{{e}^{x}}dx}=(1-(-1))\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(-1)+f\left( -1+\dfrac{2}{n} \right)+f\left( -1+2.\dfrac{2}{n} \right)+...+f\left( -1+(n-1).\dfrac{2}{n} \right) \right]$$=(1+1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}+{{e}^{\left( -1+\dfrac{2}{n} \right)}}+{{e}^{\left( -1+2.\dfrac{2}{n} \right)}}+...+{{e}^{\left( -1+(n-1).\dfrac{2}{n} \right)}} \right]$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}+{{e}^{-1}}{{e}^{\left( \dfrac{2}{n} \right)}}+{{e}^{-1}}{{e}^{\left( 2.\dfrac{2}{n} \right)}}+...+{{e}^{-1}}{{e}^{\left( (n-1).\dfrac{2}{n} \right)}} \right]$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}\left\{ 1+{{e}^{\left( \dfrac{2}{n} \right)}}+{{e}^{\left( \dfrac{4}{n} \right)}}+...+{{e}^{\left( (n-1).\dfrac{2}{n} \right)}} \right\} \right]$

It is known that the sum of n-terms in Geometric progression with $a=1\text{  ;  }r={{e}^{\dfrac{2}{n}}}$ is – $\dfrac{{{e}^{\dfrac{2n}{n}}}-1}{{{e}^{\dfrac{2}{n}}}-1}=\dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1}$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right\} \right]$

$=2{{e}^{-1}}\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left\{ \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right\}$

$=2{{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}} \right]2} \right\}$

It is known that – $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{h}}-1}{h}=1$

$=2{{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{(1)2} \right\}$

$={{e}^{-1}}({{e}^{2}}-1)$

$=({{e}^{1}}-{{e}^{-1}})$

$=e-\dfrac{1}{e}$

Thus,  $\int\limits_{-1}^{1}{{{e}^{x}}dx}=e-\dfrac{1}{e}$


6. Evaluate the definite integral as limit of sum – $\int\limits_{0}^{4}{(x+{{e}^{2x}})dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x+{{e}^{2x}}\text{ ; }a=0\text{ ; }b=4\text{ ; }h=\dfrac{4-0}{n}=\dfrac{4}{n}$

$\therefore \int\limits_{0}^{4}{(x+{{e}^{2x}})dx}=(4-0)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(0)+f\left( 0+\dfrac{4}{n} \right)+f\left( 0+2.\dfrac{4}{n} \right)+f\left( 0+3.\dfrac{4}{n} \right)+...+f\left( 0+(n-1)\dfrac{4}{n} \right) \right]$$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( 0+{{e}^{0}} \right)+\left( \dfrac{4}{n}+{{e}^{2.\dfrac{4}{n}}} \right)+\left( 2.\dfrac{4}{n}+{{e}^{2.2.\dfrac{4}{n}}} \right)+...+\left( (n-1)\dfrac{4}{n}+{{e}^{2(n-1)\dfrac{4}{n}}} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( {{e}^{0}}+{{e}^{\dfrac{8}{n}}}+{{e}^{2.\dfrac{8}{n}}}+...+{{e}^{(n-1)\dfrac{8}{n}}} \right)+\left( \dfrac{4}{n}+2.\dfrac{4}{n}+...+(n-1)\dfrac{4}{n} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( 1+{{e}^{\dfrac{8}{n}}}+{{e}^{2.\dfrac{8}{n}}}+...+{{e}^{(n-1)\dfrac{8}{n}}} \right)+\dfrac{4}{n}\left( 1+2+...+(n-1) \right) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

Also, the sum of n-terms in Geometric progression with $a=1\text{  ;  }r={{e}^{\dfrac{8}{n}}}$ is – $\dfrac{{{e}^{\dfrac{8n}{n}}}-1}{{{e}^{\dfrac{8}{n}}}-1}=\dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1}$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+\dfrac{4}{n}\left( \dfrac{n(n-1)}{2} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+2\left( n-1 \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 2\left( n-1 \right) \right]$

$=4\left\{ \dfrac{{{e}^{8}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{8}{n}}}-1}{\dfrac{8}{n}} \right]8} \right\}+4\underset{n\to \infty }{\mathop{\lim }}\,2\left[ \dfrac{n-1}{n} \right]$

$=4\left\{ \dfrac{{{e}^{8}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{8}{n}}}-1}{\dfrac{8}{n}} \right]8} \right\}+4\underset{n\to \infty }{\mathop{\lim }}\,2\left[ 1-\dfrac{1}{n} \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

And, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{h}}-1}{h}=1$

$=4\left\{ \dfrac{{{e}^{8}}-1}{(1)8} \right\}+4\left\{ 2\left[ 1-0 \right] \right\}$

$=\left\{ \dfrac{{{e}^{8}}-1}{2} \right\}+4(2)$

$=\dfrac{{{e}^{8}}-1}{2}+8$

$=\dfrac{{e^8 - 1+16}}{2}$

$=\left( \dfrac{{{e}^{8}}+15}{2} \right)$

Thus,  $\int\limits_{0}^{4}{(x+{{e}^{2x}})dx}=\dfrac{{{e}^{8}}+15}{2}$


Exercise-7.9

1. Evaluate the definite integral– $\int\limits_{-1}^{1}{(x+1)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{(x+1)dx}=\dfrac{{{x}^{2}}}{2}+x$

So, $\int\limits_{-1}^{1}{(x+1)dx=}\left[ \dfrac{{{x}^{2}}}{2}+x \right]_{-1}^{1}$

$=\left[ \dfrac{{{1}^{2}}}{2}+1 \right]-\left[ \dfrac{{{(-1)}^{2}}}{2}+(-1) \right]$

$=\left[ \dfrac{1}{2}+1 \right]-\left[ \dfrac{1}{2}-1 \right]$

$=\dfrac{1}{2}+1-\dfrac{1}{2}+1=2$

Thus, $\int\limits_{-1}^{1}{(x+1)dx=}2$


2. Evaluate the definite integral– $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\dfrac{1}{x}dx}=\log \left| x \right|$

So, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\left[ \log \left| x \right| \right]_{2}^{3}$

$=\left[ \log \left| 3 \right| \right]-\left[ \log \left| 2 \right| \right]$

$=\log \dfrac{3}{2}$

Thus, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\log \dfrac{3}{2}$


3. Evaluate the definite integral– $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=4\left( \dfrac{{{x}^{4}}}{4} \right)-5\left( \dfrac{{{x}^{3}}}{3} \right)+6\left( \dfrac{{{x}^{2}}}{2} \right)+9x$

$={{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x$

So,  $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\left[ {{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x \right]_{1}^{2}$

$=\left[ {{2}^{4}}-\dfrac{5{{(2)}^{3}}}{3}+3{{(2)}^{2}}+9(2) \right]-\left[ {{1}^{4}}-\dfrac{5{{(1)}^{3}}}{3}+3{{(1)}^{2}}+9(1) \right]$

$=\left[ 16-\dfrac{40}{3}+12+18 \right]-\left[ 1-\dfrac{5}{3}+3+9 \right]$

$=\left[ 46-\dfrac{40}{3} \right]-\left[ 13-\dfrac{5}{3} \right]$

$=46-\dfrac{40}{3}-13+\dfrac{5}{3}$

$=33-\dfrac{35}{3}$

$=\dfrac{99-35}{3}$

$=\dfrac{64}{3}$

Thus, $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\dfrac{64}{3}$


4. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\sin 2xdx}=\dfrac{-\cos 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\left[ \dfrac{-\cos 2x}{2} \right]_{0}^{\dfrac{\pi }{4}}$

$=\left[ \dfrac{-\cos 2\left( \dfrac{\pi }{4} \right)}{2} \right]-\left[ \dfrac{-\cos 0}{2} \right]$

$=\left[ \dfrac{-\cos \left( \dfrac{\pi }{2} \right)}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\left[ \dfrac{0}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\dfrac{1}{2}$

Thus, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\dfrac{1}{2}$


5. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\cos 2xdx}=\dfrac{\sin 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=\left[ \dfrac{\sin 2x}{2} \right]_{0}^{\dfrac{\pi }{2}}$

$=\left[ \dfrac{\sin 2\left( \dfrac{\pi }{2} \right)}{2} \right]-\left[ \dfrac{\sin 0}{2} \right]$

$=\left[ \dfrac{\sin \pi }{2} \right]+\left[ \dfrac{0}{2} \right]$

$=0+0$

$=0$

Thus,  $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=0$


6. Evaluate the definite integral– $\int\limits_{4}^{5}{{{e}^{x}}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{{{e}^{x}}dx}={{e}^{x}}$

So, $\int\limits_{4}^{5}{{{e}^{x}}dx}=\left[ {{e}^{x}} \right]_{4}^{5}$

$=\left[ {{e}^{5}} \right]-\left[ {{e}^{4}} \right]$

$={{e}^{4}}(e-1)$

Thus, $\int\limits_{4}^{5}{{{e}^{x}}dx}={{e}^{4}}(e-1)$


7. $\int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}$

Ans: We know that,

$\int{\tan xdx}=-\log \left| \cos x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int{\tan x dx} = \left[ -\log \left| \cos x \right| \right]_{0}^{\dfrac{\pi}{4}}$

(image will be uploaded soon)

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \cos \dfrac{\pi }{4} \right|+\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \dfrac{1}{\sqrt{2}} \right|+\log \left| 1 \right| \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\dfrac{1}{2}\log 2$


8. $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}$

Ans: We know that,

$\int{\cos ecxdx}=\log \left| \cos ecx-\cot x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ecx-\cot x \right| \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ec\dfrac{\pi }{4}-\cot \dfrac{\pi }{4} \right|-\log \left| \cos ec\dfrac{\pi }{6}-\cot \dfrac{\pi }{6} \right| \right]$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \sqrt{2}-1 \right|-\log \left| 2-\sqrt{3} \right| \right]$

$\therefore \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\log \left( \dfrac{\sqrt{2}-1}{2-\sqrt{3}} \right)$


9. $\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}$

Ans: We know that,

$\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}={{\sin }^{-1}}x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}1-{{\sin }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ \dfrac{\pi }{2}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\dfrac{\pi }{2}$


10. $\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}$

Ans: We know that,

$\int{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ \dfrac{\pi }{4}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{4}$


11. $\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}$

Ans: We know that,

$\int{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\log \left| \dfrac{x-1}{x+1} \right|+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{x-1}{x+1} \right| \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{3-1}{3+1} \right|-\log \left| \dfrac{2-1}{2+1} \right| \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \dfrac{1}{2}-\log \dfrac{1}{3} \right]$

$\therefore \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\log \dfrac{3}{2}$


12. $\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}$

Ans: We know that,

$\int{{{\cos }^{2}}xdx}=\int{\left( \dfrac{1+\cos 2x}{2} \right)dx}$

$\Rightarrow \int{{{\cos }^{2}}xdx}=\dfrac{x}{2}+\dfrac{\sin 2x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\left[ \dfrac{x}{2}+\dfrac{\sin 2x}{4} \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-\dfrac{\sin \pi }{2}-0-\dfrac{\sin 0}{2} \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}+0-0-0 \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$


13. $\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}$

Ans: We know that,

$\int{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\int{\left( \dfrac{2x}{{{x}^{2}}+1} \right)dx}$

$\Rightarrow \int{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)$

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\left[ \log \left( 1+{{3}^{2}} \right)-\log \left( 1+{{2}^{2}} \right) \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\left[ \log 10-\log 5 \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \dfrac{10}{5}$

$\therefore \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{4}\log 2$


14. $\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$

Ans: Solving $\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$ ,

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{5\left( 2x+3 \right)}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x+15}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x}{5{{x}^{2}}+1}dx}+3\int{\dfrac{1}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x}{5{{x}^{2}}+1}dx}+3\int{\dfrac{1}{5\left( {{x}^{2}}+\dfrac{1}{5} \right)}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\log \left( 5{{x}^{2}}+1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\left( \sqrt{5} \right)x$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\left\{ \dfrac{1}{5}\log \left( 5+1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\left( \sqrt{5} \right)x \right\}-\left\{ \dfrac{1}{5}\log \left( 1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}0 \right\}$

$\therefore \int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\log 6+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\sqrt{5}$


15. $\int\limits_{0}^{1}{x{{e}^{{{x}^{2}}}}dx}$

Ans: Let ${{x}^{2}}=t$ ,

Differentiating it we get,

$2xdx=dt$

Therefore, the integral becomes,

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\left[ \dfrac{1}{2}{{e}^{t}} \right]_{0}^{1}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\dfrac{1}{2}e-\dfrac{1}{2}{{e}^{0}}$

$\dfrac{1}{2}\left( e-1 \right)$


16. $\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}$

Ans: The given integral can be written as

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\int\limits_{1}^{2}{\left\{ 5-\dfrac{20x+15}{{{x}^{2}}+4x+3} \right\}dx}$

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\left[ 5x \right]_{1}^{2}-\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$                             …(1)

Solving $\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$ ,

Let $20x+15=A\dfrac{d}{dx}\left( {{x}^{2}}+4x+3 \right)+B$

Equating the coefficients of $x$ and constant term we get,

$A=10,B=-25$

Let ${{x}^{2}}+4x+3=t$

Differentiating it we get,

$\left( 2x+4 \right)dx=dt$

Therefore, the integral becomes

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}$

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}=10\log t-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+2-1}{x+2+1} \right) \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\left[ 10\log \left( {{x}^{2}}+4x+3 \right)-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+1}{x+3} \right) \right] \right]_{1}^{2}$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 15-10\log 8-25\left[ \dfrac{1}{2}\log \dfrac{3}{5}-\dfrac{1}{2}\log \dfrac{2}{4} \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 5+10\log 3-10\log 4-10\log 2-\dfrac{25}{2}\left[ \log 3-\log 5-\log 2+\log 4 \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log 5-\dfrac{45}{2}\log 4-\dfrac{5}{2}\log 3+\dfrac{5}{2}\log 2$

$\therefore \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2}$

Substituting it in (1) we get,

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\left[ \dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2} \right]$

$\therefore \int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\dfrac{5}{2}\left[ 9\log \dfrac{5}{4}-\log \dfrac{3}{2} \right]$


17. $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}$

Ans: We know that,

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan x+\dfrac{{{x}^{4}}}{4}+2x$

Therefore, by second fundamental theorem of calculus

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan x+\dfrac{{{x}^{4}}}{4}+2x \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan \dfrac{\pi }{4}+\dfrac{1}{4}{{\left( \dfrac{\pi }{4} \right)}^{2}}+2\left( \dfrac{\pi }{4} \right)-\left( 2\tan 0+0+0 \right) \right]$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan \dfrac{\pi }{4}+\dfrac{{{\pi }^{4}}}{{{4}^{5}}}+\dfrac{\pi }{2}$

$\therefore \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2+\dfrac{\pi }{2}+\dfrac{{{\pi }^{4}}}{1024}$


18. $\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}$

Ans: We know that,

$\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}$

$\Rightarrow -\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\cos xdx}$

$\int{\cos xdx}=\sin x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\pi }{\cos xdx}=\sin \pi -sin0$

$\therefore \int\limits_{0}^{\pi }{\cos xdx}=0$


19. $\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}$

Ans: Solving the integral we get,

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x+1}{{{x}^{2}}+4}dx}$

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x}{{{x}^{2}}+4}dx}+3\int{\dfrac{1}{{{x}^{2}}+4}dx}$

$\therefore \int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2} \right]_{0}^{2}$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{2}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{2}{2}-3\log \left( {{0}^{2}}+4 \right)-\dfrac{3}{2}{{\tan }^{-1}}\dfrac{0}{2} \right]$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}{{\tan }^{-1}}1-3\log 4-\dfrac{3}{2}{{\tan }^{-1}}0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}\left( \dfrac{\pi }{4} \right)-3\log 4-0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \dfrac{8}{4}+\dfrac{3\pi }{8}$

$\therefore \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 2+\dfrac{3\pi }{8}$


20. $\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}$

Ans: Solving the integral we get,

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x\int{{{e}^{x}}dx}-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{{{e}^{x}}dx} \right\}dx}+\left\{ \dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi }{4}} \right\}$

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-\int{{{e}^{x}}dx}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-{{e}^{x}}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1{{e}^{1}}-{{e}^{1}}-\dfrac{4}{\pi }\cos \dfrac{\pi }{4} \right)-\left( 0{{e}^{0}}-{{e}^{0}}-\dfrac{4}{\pi }\cos 0 \right)$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1+\dfrac{4}{\pi }-\dfrac{2\sqrt{2}}{\pi } \right)$


21. $\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}$

  1. $\dfrac{\pi }{3}$

  2. $\dfrac{2\pi }{3}$

  3. $\dfrac{\pi }{6}$

  4. $\dfrac{\pi }{12}$

Ans: Solving the integral we get,

$\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x$

Therefore, by second fundamental theorem of calculus

$\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}1$

$\Rightarrow \int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{3}-\dfrac{\pi }{4}$

$\therefore \int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{12}$

Thus, the correct option is (D)


22. $\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}$

  1. $\dfrac{\pi }{6}$

  2. $\dfrac{\pi }{12}$

  3. $\dfrac{\pi }{24}$

  4. $\dfrac{\pi }{4}$

Ans: Solving the integral we get,

$\int{\dfrac{1}{4+9{{x}^{2}}}dx}=\int{\dfrac{1}{{{2}^{2}}+{{\left( 3x \right)}^{2}}}dx}$

Let $3x=t$ ,

Differentiating it we get,

$3dx=dt$

$\therefore \int{\dfrac{1}{{{2}^{2}}+{{\left( 3x \right)}^{2}}}dx}=\dfrac{1}{6}{{\tan }^{-1}}\left( \dfrac{3x}{2} \right)$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}{{\tan }^{-1}}\left( \dfrac{3}{2}\cdot \dfrac{2}{3} \right)-\dfrac{1}{6}{{\tan }^{-1}}0$

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}\cdot \dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{\pi }{24}$

Thus, the correct answer is (C).


Exercise 7.10

Solve the following integrals.

1. $\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}$

Ans: Let ${{x}^{2}}+1$

Differentiating $2xdx=dt$ ,

Therefore, the integral becomes

$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{dt}{t}}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log \left| t \right| \right]_{1}^{2}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log 2-\log 1 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\log 2$


2. $\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }$

Ans: The integral can be written as:

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{4}}\phi \cos \phi d\phi }$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$

Let $\sin \phi =t$

Differentiating it we get,

$\cos \phi d\phi =dt$

Therefore, the integral becomes

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}{{\left( 1-{{t}^{2}} \right)}^{2}}dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}\left( 1+{{t}^{4}}-2{{t}^{2}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\left( {{t}^{\dfrac{1}{2}}}+{{t}^{\dfrac{9}{2}}}-2{{t}^{\dfrac{5}{2}}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\left[ \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{{{t}^{\dfrac{11}{2}}}}{\dfrac{11}{2}}-\dfrac{2{{t}^{\dfrac{7}{2}}}}{\dfrac{7}{2}} \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{2}{3}+\dfrac{2}{11}-\dfrac{4}{7}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{154+42-132}{231}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$


3. $\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}$

Ans: Let $x=\tan \theta $

Differentiating it we get,

$dx={{\sec }^{2}}\theta d\theta $

Therefore, the integral becomes

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\left( \tan \theta  \right)}^{2}}} \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \sin 2\theta  \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\int\limits_{0}^{\dfrac{\pi }{4}}{\theta {{\sec }^{2}}\theta d\theta }$

Let $u=\theta $

And $v={{\sec }^{2}}\theta $

Using integration by parts we get,

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left\{ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right\}d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\int{\tan \theta d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\log \left| \cos \theta  \right| \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}\tan \dfrac{\pi }{4}-\log \left| \cos \dfrac{\pi }{4} \right|-\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}-\log \left| \dfrac{1}{\sqrt{2}} \right|-\log 1 \right]$

$\therefore \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\dfrac{\pi }{2}-\log 2$


4. $\int\limits_{0}^{2}{x\sqrt{x+2}dx}$ 

(Put $x+2={{t}^{2}}$)

Ans: Let $x+2={{t}^{2}}$

Differentiating it, $dx=2tdt$

Therefore, the integral becomes

$\int\limits_{0}^{2}{x\sqrt{x+2}dx}=\int\limits_{\sqrt{2}}^{2}{2{{t}^{2}}\left( {{t}^{2}}-2 \right)dt}$

$\Rightarrow \int\limits_{0}^{2}{x\sqrt{x+2}dx}=2\left[ \dfrac{{{t}^{5}}}{5}-\dfrac{2{{t}^{3}}}{3} \right]_{\sqrt{2}}^{2}$

$\Rightarrow \int\limits_{0}^{2}{x\sqrt{x+2}dx}=2\left[ \dfrac{35}{5}-\dfrac{16}{3}-\dfrac{4\sqrt{2}}{5}+\dfrac{4\sqrt{2}}{3} \right]$

$\therefore \int\limits_{0}^{2}{x\sqrt{x+2}dx}=\dfrac{16\sqrt{2}\left( \sqrt{2}+1 \right)}{15}$


5. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}$

Ans: Let $\cos x=t$

Differentiating it, $-\sin xdx=dt$

Therefore, the integral becomes

$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\int\limits_{1}^{0}{\dfrac{dt}{1+{{t}^{2}}}}$

$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\left[ {{\tan }^{-1}}t \right]_{1}^{0}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\left[ {{\tan }^{-1}}0-{{\tan }^{-1}}1 \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=\dfrac{\pi }{4}$


6. $\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}$

Ans: The integral can be written as

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{0}^{2}{\dfrac{dx}{-\left( {{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{17}{4} \right)}}$

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{0}^{2}{\dfrac{dx}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( x-\dfrac{1}{2} \right)}^{2}} \right)}}$

Let $x-\dfrac{1}{2}=t$

Differentiating it we get,

$dx=dt$

Therefore, the integral becomes

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\left[ \dfrac{1}{2\left( \dfrac{\sqrt{17}}{2} \right)}\log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)+t}{\left( \dfrac{\sqrt{17}}{2} \right)-t} \right]_{\dfrac{-1}{2}}^{\dfrac{3}{2}}$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)+\dfrac{3}{2}}{\left( \dfrac{\sqrt{17}}{2} \right)-\dfrac{3}{2}}-\log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)-\dfrac{1}{2}}{\left( \dfrac{\sqrt{17}}{2} \right)+\dfrac{1}{2}} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}-\log \dfrac{\sqrt{17}-1}{\sqrt{17}+1} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}\times \dfrac{\sqrt{17}+1}{\sqrt{17}-1} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{25+17+10\sqrt{17}}{8} \right]$

$\therefore \int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{21+5\sqrt{17}}{4} \right]$


7. $\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}$

Ans: The integral can be written as

$\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\int\limits_{-1}^{1}{\dfrac{1}{{{\left( x-1 \right)}^{2}}+{{2}^{2}}}dx}$

Let $x+1=t$

Differentiating it we get,

$dx=dt$

Therefore, the integral becomes

$\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\int\limits_{0}^{2}{\dfrac{dt}{{{t}^{2}}+{{2}^{2}}}}$

$\Rightarrow \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\left[ \dfrac{1}{2}{{\tan }^{-1}}\dfrac{t}{2} \right]_{0}^{2}$

$\Rightarrow \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\left[ \dfrac{1}{2}{{\tan }^{-1}}1-\dfrac{1}{2}{{\tan }^{-1}}0 \right]$

$\therefore \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\dfrac{\pi }{8}$


8. $\int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}$

Ans: Let $2x=t$

Differentiating it $2dx=dt$ ,

Therefore, the integral becomes

$\int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\dfrac{1}{2}\int\limits_{2}^{1}{\left( \dfrac{2}{t}-\dfrac{2}{{{t}^{2}}} \right){{e}^{t}}dt}$

$\Rightarrow \int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}$

Let $\dfrac{1}{t}=f\left( t \right)$

Then, $f'\left( t \right)=-\dfrac{1}{{{t}^{2}}}$

$\int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\int\limits_{2}^{1}{\left[ f\left( t \right)+f'\left( t \right) \right]{{e}^{t}}dt}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\left[ {{e}^{t}}f\left( t \right) \right]_{2}^{4}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\left[ \dfrac{{{e}^{t}}}{t} \right]_{2}^{4}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\dfrac{{{e}^{4}}}{4}-\dfrac{{{e}^{2}}}{2}$

$\therefore \int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\dfrac{{{e}^{2}}\left( {{e}^{2}}-2 \right)}{4}$


9. The value of the integral $\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}$

  1. $6$

  2. $0$

  3. $3$

  4. $4$

Ans: Let $x=\sin \theta $

Differentiating it, $dx=\cos \theta d\theta $

Therefore, the integral becomes

$\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta  \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta  \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int   \limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta  \right)}^{\dfrac{1}{3}}}{{\left( \cos \theta  \right)}^{\dfrac{2}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta  \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \cos \theta  \right)}^{\dfrac{5}{3}}}}{{{\sin }^{\dfrac{5}{3}}}\theta }\cos e{{c}^{2}}\theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta  \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{{{\left( \cot \theta  \right)}^{\dfrac{5}{3}}}\cos e{{c}^{2}}\theta d\theta }$

Let $\cot \theta =t$

Differentiating it, $-\cos e{{c}^{2}}\theta d\theta =dt$

Therefore, the integral becomes

$\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\int\limits_{2\sqrt{2}}^{0}{{{t}^{\dfrac{5}{3}}}dt}$

$\Rightarrow \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\dfrac{3}{8}{{\sqrt{8}}^{\dfrac{8}{3}}}$

$\Rightarrow \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\dfrac{3}{8}\times 16$

$\therefore \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=6$

Thus, the correct answer is A.


10. If $f\left( x \right)=\int\limits_{0}^{x}{t\sin tdt}$ ,then $f'\left( x \right)$ is

  1. $\cos x+x\sin x$

  2. $x\sin x$

  3. $x\cos x$

  4. $\sin x+x\cos x$

Ans: Given $f\left( x \right)=\int\limits_{0}^{x}{t\sin tdt}$

Using Integration by parts, we get

$f\left( x \right)=t\int\limits_{0}^{x}{\sin tdt}-\int\limits_{0}^{x}{\left\{ \left( \dfrac{d}{dt}t \right)\int{\sin tdt} \right\}dt}$

$\Rightarrow f\left( x \right)=\left[ t\left( -\cos t \right) \right]_{0}^{x}-\int\limits_{0}^{x}{-\cot tdt}$

$\Rightarrow f\left( x \right)=\left[ -t\left( \cos t \right)+\sin t \right]_{0}^{x}$

$\Rightarrow f\left( x \right)=-x\cos x+\sin x$

Therefore,

$f'\left( x \right)=-\left[ -x\sin x+\cos x \right]+\cos x$

$\therefore f'\left( x \right)=x\sin x$

Thus, the correct answer is B.


Exercise 7.11

Solve the following integrals.

1. $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$                          …(2)

Adding equation (1) and (2),

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$


2. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}=\dfrac{\pi }{4}$


3. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}=\dfrac{\pi }{4}$


4. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{5}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}x+{{\cos }^{5}}x}{{{\sin }^{5}}x+{{\cos }^{5}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}=\dfrac{\pi }{4}$


5. $\int\limits_{-5}^{5}{\left| x+2 \right|}dx$

Ans:  Let $I=\int\limits_{-5}^{5}{\left| x+2 \right|}dx$ 

Since, $\left( x+2 \right)\le 0$ for interval $\left[ -5,-2 \right]$ .

Therefore, $\left( x+2 \right)\ge 0$ for interval $\left[ -2,5 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$ .

Thus, 

$I=\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-5}^{2}+\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-2}^{5}$

$=-\left[ \dfrac{{{\left( 2 \right)}^{2}}}{2}+2\left( 2 \right)-\dfrac{{{\left( -5 \right)}^{2}}}{2}-2\left( -5 \right) \right]+\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}+2\left( 5 \right)-\dfrac{{{\left( -2 \right)}^{2}}}{2}-2\left( -2 \right) \right]$$=-\left[ 2-4-\dfrac{25}{2}+10 \right]+\left[ \dfrac{25}{2}+10-2+4 \right]$

$=-2+4+\dfrac{25}{2}+10+\dfrac{25}{2}+10-2+4$

$=29$


6. $\int\limits_{2}^{8}{\left| x-5 \right|}dx$

Ans: Let $I=\int\limits_{2}^{8}{\left| x-5 \right|}dx$ 

Since, $\left( x-5 \right)\le 0$ for interval $\left[ 2,5 \right]$ .

Therefore, $\left( x-5 \right)\ge 0$ for interval $\left[ 5,8 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$ .

Thus, 

$I=\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{2}^{5}+\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{5}^{8}$

$=-\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}-5\left( 5 \right)-\dfrac{{{\left( 2 \right)}^{2}}}{2}+5\left( 2 \right) \right]+\left[ \dfrac{{{\left( 8 \right)}^{2}}}{2}-5\left( 8 \right)-\dfrac{{{\left( 5 \right)}^{2}}}{2}+5\left( 5 \right) \right]$

$=-\left[ \dfrac{25}{2}-25-2+10 \right]+\left[ 32-40-\dfrac{25}{2}+25 \right]$

$=-\dfrac{25}{2}+25+2-10+32-40-\dfrac{25}{2}+25$

$=9$


7. $\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}$

Ans: Let  $I=\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}$

Thus, $I=\int\limits_{0}^{1}{\left( 1-x \right){{\left( 1-\left( 1-x \right) \right)}^{n}}dx}$

Since, $\int\limits_{1}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$.

Therefore, 

$\int\limits_{0}^{1}{\left( 1-x \right){{\left( x \right)}^{n}}}dx$

$=\int\limits_{0}^{1}{\left( {{x}^{n}}-{{x}^{n+1}} \right)dx}$

$=\left[ \dfrac{{{x}^{n+1}}}{n+1}-\dfrac{{{x}^{n+2}}}{n+2} \right]_{0}^{1}$

$=\left[ \dfrac{1}{n+1}-\dfrac{1}{n+2} \right]$

$=\dfrac{\left( n+2 \right)-\left( n+1 \right)}{\left( n+1 \right)\left( n+2 \right)}$

$=\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}$


8. $\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$

Ans:  Let  $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\tan \left( \dfrac{\pi }{4}-x \right) \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x} \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left\{ 1+\dfrac{1-\tan x}{1+\tan x} \right\}dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \dfrac{2}{1+\tan x}}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-I$

$2I=\left[ x\log 2 \right]_{0}^{\dfrac{\pi }{4}}$

$2I=\dfrac{\pi }{4}\log 2$

$I=\dfrac{\pi }{8}\log 2$


9. $\int\limits_{0}^{2}{x\sqrt{2-x}dx}$

Ans: Let  $I=\int\limits_{0}^{2}{x\sqrt{2-x}dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)}dx=\int\limits_{0}^{a}{f\left( a-x \right)}dx$

Therefore, $I=\int\limits_{0}^{2}{\left( 2-x \right)}\sqrt{x}dx$

$=\int\limits_{0}^{2}{\left\{ 2{{x}^{{1}/{2}\;}}-{{x}^{{3}/{2}\;}} \right\}}dx$

$=\left[ 2\left( \dfrac{{{x}^{{3}/{2}\;}}}{{3}/{2}\;} \right)-\dfrac{{{x}^{{5}/{2}\;}}}{{5}/{2}\;} \right]_{0}^{2}$

$=\left[ \dfrac{4}{3}{{x}^{{3}/{2}\;}}-\dfrac{2}{5}{{x}^{{5}/{2}\;}} \right]_{0}^{2}$

$=\dfrac{4}{3}{{\left( 2 \right)}^{{3}/{2}\;}}-\dfrac{2}{5}{{\left( 2 \right)}^{{5}/{2}\;}}$

$=\dfrac{8\sqrt{2}}{3}-\dfrac{8\sqrt{2}}{5}$

$=\dfrac{40\sqrt{2}-24\sqrt{2}}{15}$

$=\dfrac{16\sqrt{2}}{15}$


10. $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}$

Ans: Let  $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \left( 2\sin x\cos x \right) \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin x-\log \cos x-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x-\log \cos x-\log 2 \right)dx}$                      …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \cos x-\log \sin x-\log 2 \right)dx}$ …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\log 2-\log 2 \right)dx}$

$\Rightarrow 2I=-2\log 2\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$

$\Rightarrow I=-\log 2\left[ \dfrac{\pi }{2} \right]$

$\Rightarrow I=-\dfrac{\pi }{2}\left[ \log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ -\log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ \log \dfrac{1}{2} \right]$

$\Rightarrow I=\dfrac{\pi }{2}\log \dfrac{1}{2}$


11. $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

Ans: Let  $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

Since, ${{\sin }^{2}}x$ is an even function.

Therefore , $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}=2\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

As if $f\left( x \right)$ is an even function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$.

Hence,

$I=2\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

$=2\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1-\cos 2x}{2}dx}$

$=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1-\cos 2x \right)dx}$

$=\left[ x-\dfrac{sin2x}{2} \right]_{0}^{\dfrac{\pi }{2}}$

$=\dfrac{\pi }{2}$


12. $\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$

Ans: Let  $I=\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$               …….(1)

Since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, $I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)}{1+\sin x\left( \pi -x \right)}dx}$

$I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)}{1+\sin x}dx}$                   …..(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\pi }{\dfrac{\pi }{1+\sin x}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)+\left( 1-\sin x \right)}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\dfrac{\left( 1-\sin x \right)}{{{\cos }^{2}}x}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\left\{ {{\sec }^{2}}x-\tan x\sec x \right\}dx}$

$\Rightarrow 2I=\pi \left[ 2 \right]$

$\Rightarrow I=\pi $


13. $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{7}}xdx}$.

Ans: Let  $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{7}}xdx}$

Since, ${{\sin }^{7}}x$ is an even function.

Therefore , $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}=0$

As if $f\left( x \right)$ is an odd function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$.

Hence, $I=0$


14. $\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$.

Ans: Let  

$I=\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$      

${{\cos }^{5}}\left( 2\pi -x \right)={{\cos }^{5}}x$                   …..(1)

If $f\left( 2a-x \right)=f\left( x \right)$ then $\int\limits_{0}^{2a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$.

If $f\left( 2a-x \right)=-f\left( x \right)$ then $\int\limits_{0}^{2a}{f\left( x \right)dx}=0$

Since, ${{\cos }^{5}}\left( \pi -x \right)=-{{\cos }^{5}}x$

Therefore, 

$I=2\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$

$\Rightarrow I=2\left( 0 \right)$

$\Rightarrow I=0$


15. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\sin x\cos x}dx}$

Ans:   Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\sin x\cos x}dx}$                     ……(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( \dfrac{\pi }{2}-x \right)-\cos \left( \dfrac{\pi }{2}-x \right)}{1+\sin \left( \dfrac{\pi }{2}-x \right)\cos \left( \dfrac{\pi }{2}-x \right)}dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\cos x\sin x}dx}$                               ……(2)

On adding equation 1 and 2

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{0}{1+\cos x\sin x}dx}$

$\Rightarrow I=0$


16. $\int\limits_{0}^{\pi }{\log \left( 1+\cos x \right)dx}$

Ans: Let  

$I=\int\limits_{0}^{\pi }{\log \left( 1+\cos x \right)dx}$       …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \left( 1+\cos \left( \pi -x \right) \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \left( 1-\cos x \right)dx}$                 …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\pi }{\left\{ \log \left( 1-\cos x \right)+\log \left( 1-\cos x \right) \right\}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\pi }{\log \left( 1-{{\cos }^{2}}x \right)dx}$

 $\Rightarrow 2I=\int\limits_{0}^{\pi }{\log {{\sin }^{2}}xdx}$

$\Rightarrow 2I=2\int\limits_{0}^{\pi }{\log \sin xdx}$

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \sin xdx}$                                     …..(3)

Since, $\sin \left( \pi -x \right)=\sin x$

Therefore, $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \sin xdx}$                        ……(4)

$\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\log \sin \left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cos xdx}$                                   …….(5)

On adding equation 4 and 5-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x+\log \cos x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x+\log \cos x+\log 2-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log 2\sin x\cos x-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log 2\sin x\cos x \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{\log 2dx}$

Let $2x=t$

On differentiating-

$2dx=dt$

If $x=0$ then $t=0$.

Thus,

$\Rightarrow I=\dfrac{I}{2}-\dfrac{\pi }{2}\log 2$

$\Rightarrow \dfrac{I}{2}=-\dfrac{\pi }{2}\log 2$

$\Rightarrow I=-\pi \log 2$


17. $\int\limits_{0}^{a}{\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx}$

Ans:  Let $I=\int\limits_{0}^{a}{\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx}$                              …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$I=\int\limits_{0}^{a}{\dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx}$                                     …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{a}{\dfrac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx}$

$\Rightarrow 2I=\int\limits_{0}^{a}{dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{a}$

$\Rightarrow 2I=a$

$\Rightarrow I=\dfrac{a}{2}$


18. $\int\limits_{0}^{4}{\left| x-1 \right|dx}$

Ans: Let $I=\int\limits_{0}^{4}{\left| x-1 \right|dx}$

Thus, $\left( x-1 \right)\le 0$ when $0\le x\le 1$  and $\left( x-1 \right)\ge 0$ when $1\le x\le 4$

Since, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)}$

Therefore, $I=\int\limits_{0}^{1}{\left| x-1 \right|}dx+\int\limits_{1}^{4}{\left| x-1 \right|dx}$

$\Rightarrow I=\int\limits_{0}^{1}{-\left( x-1 \right)}dx+\int\limits_{1}^{4}{\left( x-1 \right)dx}$

$=\left[ x-\dfrac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{1}^{4}$

$=1-\dfrac{1}{2}+\dfrac{{{\left( 4 \right)}^{2}}}{2}-4-\dfrac{1}{2}+1$

$=1-\dfrac{1}{2}+8-4-\dfrac{1}{2}+1$

$=5$


19. Show that $\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)}dx$ , if $f$and $g$ are defined as $f\left( x \right)=f\left( a-x \right)$ and $g\left( x \right)+g\left( a-x \right)=4$ .

Ans: Let $\int\limits_{0}^{a}{f\left( x \right)}g\left( x \right)dx$              …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow \int\limits_{0}^{a}{f\left( a-x \right)}g\left( a-x \right)dx$

$\Rightarrow \int\limits_{0}^{a}{f\left( x \right)}g\left( a-x \right)dx$   ……(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{a}{\left\{ f\left( x \right)g\left( x \right)+f\left( x \right)g\left( a-x \right) \right\}dx}$

$\Rightarrow 2I=\int\limits_{0}^{a}{f\left( x \right)\left\{ g\left( x \right)+g\left( a-x \right) \right\}dx}$

As, $g\left( x \right)+g\left( a-x \right)=4$.

Thus,

 $\Rightarrow 2I=\int\limits_{0}^{a}{4f\left( x \right)dx}$

$\Rightarrow I=2\int\limits_{0}^{a}{f\left( x \right)dx}$

Hence, $\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)}dx$ , if $f$and $g$ are defined as $f\left( x \right)=f\left( a-x \right)$ and $g\left( x \right)+g\left( a-x \right)=4$ .


20. The value of $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}}+x\cos x+{{\tan }^{5}}x+1 \right)dx}$ is 

  1. $0$

  2. $2$

  3. $\pi $

  4. $1$

Ans: Let $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}}+x\cos x+{{\tan }^{5}}x+1 \right)dx}$

$\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}} \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( x\cos x \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{\tan }^{5}}x \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( 1 \right)dx}$

If $f\left( x \right)$ is an even function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)}dx$

And $f\left( x \right)$ is an odd function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$

Thus,

$I=0+0+0+2\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$

$=2\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$=2\left[ \dfrac{\pi }{2} \right]$

$=\pi $

Hence, the correct option is C.


21. The value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$ is

  1. $2$

  2. $\dfrac{3}{4}$

  3. $0$

  4. $-2$

Ans: Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$                …..(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin \left( \dfrac{\pi }{2}-x \right)}{4+3\cos \left( \dfrac{\pi }{2}-x \right)} \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\cos x}{4+3\sin x} \right)dx}$                       …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log 1dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{0dx}$

$\Rightarrow I=0$

Hence, the correct option is C.


Miscellaneous Exercise

1. $\dfrac{1}{x-{{x}^{3}}}$

Ans: Given $\dfrac{1}{x-{{x}^{3}}}$

So, $\dfrac{1}{x-{{x}^{3}}}=\dfrac{1}{x\left( 1-{{x}^{2}} \right)}$

$=\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}$

Let $\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{A}{x}+\dfrac{B}{\left( 1-x \right)}+\dfrac{C}{1+x}$                 …….(1)

$\Rightarrow 1=A\left( 1-{{x}^{2}} \right)+Bx\left( 1+x \right)+Cx\left( 1-x \right)$

$\Rightarrow 1=A-A{{x}^{2}}+Bx+B{{x}^{2}}+Cx-C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},x$ and constant term –

$-A+B-C=0$              

$B+C=0$

$A=1$

Thus, we get $A=1,B=\dfrac{1}{2}$ and $C=-\dfrac{1}{2}$

By equation 1-

$\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{1}{x}+\dfrac{1}{2\left( 1-x \right)}-\dfrac{1}{2\left( 1+x \right)}$

$\Rightarrow \int{\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}}dx=\int{\dfrac{1}{x}dx}+\dfrac{1}{2}\int{\dfrac{1}{\left( 1-x \right)}}dx-\dfrac{1}{2}\int{\dfrac{1}{\left( 1+x \right)}dx}$

$=\log x-\dfrac{1}{2}\log \left( 1-x \right)-\dfrac{1}{2}\log \left( 1+x \right)$

$=\log x-\dfrac{1}{2}\log {{\left( 1-x \right)}^{\dfrac{1}{2}}}-\dfrac{1}{2}\log {{\left( 1+x \right)}^{\dfrac{1}{2}}}$

$=\log \left( \dfrac{x}{{{\left( 1-x \right)}^{\dfrac{1}{2}}}{{\left( 1+x \right)}^{\dfrac{1}{2}}}} \right)+C$

$=\log {{\left( \dfrac{{{x}^{2}}}{\left( 1-{{x}^{2}} \right)} \right)}^{\dfrac{1}{2}}}+C$

$=\dfrac{1}{2}\log \left( \dfrac{{{x}^{2}}}{1-{{x}^{2}}} \right)+C$


2. $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$

Ans: Given $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$

$=\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}\times \dfrac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{\left( x+a \right)-\left( x+b \right)}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{a-b}$

$\Rightarrow \int{\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}}dx=\dfrac{1}{a-b}\int{\left( \sqrt{x+a}-\sqrt{x+b} \right)}dx$

$=\dfrac{1}{a-b}\left[ \dfrac{{{\left( x+a \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{\left( x+b \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]$

$=\dfrac{2}{3\left( a-b \right)}\left[ {{\left( x+a \right)}^{\dfrac{3}{2}}}-{{\left( x+b \right)}^{\dfrac{3}{2}}} \right]+C$


3. $\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}$                    $\left[ \text{Hint: x=}\dfrac{a}{t} \right]$

Ans: Given

$\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}$

Let $x=\dfrac{a}{t}$

On differentiating-

$dx=-\dfrac{a}{{{t}^{2}}}dt$

$\Rightarrow \int{\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}dx=\int{\dfrac{1}{\dfrac{a}{t}\sqrt{a.\dfrac{a}{t}-{{\left( \dfrac{a}{t} \right)}^{2}}}}}\left( -\dfrac{a}{{{t}^{2}}}dt \right)}$

$=-\int{\dfrac{1}{at}.\dfrac{1}{\sqrt{\dfrac{1}{t}-\dfrac{1}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{\dfrac{{{t}^{2}}}{t}-\dfrac{{{t}^{2}}}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{t-1}}}dt$

$=-\dfrac{1}{a}\left[ 2\sqrt{t-1} \right]+C$

Substituting value of t-

$=-\dfrac{1}{a}\left[ 2\sqrt{\dfrac{a}{x}-1} \right]+C$

$=-\dfrac{2}{a}\left[ \dfrac{\sqrt{a-x}}{\sqrt{x}} \right]+C$

$=-\dfrac{2}{a}\left[ \sqrt{\dfrac{a-x}{x}} \right]+C$


4. Integrate the expression: $\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}{{\left( {{\text{x}}^{\text{4}}}\text{+1} \right)}^{\dfrac{\text{3}}{\text{4}}}}}$. 

Ans:The given expression is, $\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}$. 

On multiplying and dividing by ${{x}^{-3}}$, the following can be obtained as shown below,

$\dfrac{{{x}^{-3}}}{{{x}^{2}}.{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}=\dfrac{{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{2}}-{{x}^{-3}}}=\dfrac{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{5}}{{\left( {{x}^{4}} \right)}^{\dfrac{-3}{4}}}}=\dfrac{1}{{{x}^{5}}}{{\left( \dfrac{{{x}^{4}}+1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}=\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}$ 

Now, consider $\dfrac{1}{{{x}^{4}}}=t$ 

$\therefore \dfrac{-4}{{{x}^{5}}}=\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dx}{{{x}^{5}}}=\dfrac{-dt}{4}$

$\therefore \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\int{\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}dx}=\dfrac{-1}{4}\int{{{\left( 1+t \right)}^{\dfrac{-3}{4}}}dt}$ 

$\Rightarrow \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\dfrac{-1}{4}\left[ \dfrac{{{\left( 1+t \right)}^{\dfrac{1}{4}}}}{\dfrac{1}{4}} \right]+C=-{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{1}{4}}}+C$, where $C$ is any arbitrary constant.


5. Integrate the expression: $\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\text{+}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}}$. $\text{[}$Hint: $\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\text{+}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}\left( \text{1+}{{\text{x}}^{\dfrac{\text{1}}{\text{6}}}} \right)}$ Put, $\text{x=}{{\text{t}}^{\text{6}}}$ $\text{]}$

Ans: The given expression is, $\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}$. 

Observe the given hint and obtain as shown below,

$\therefore \dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}=\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}$ 

Consider $x={{t}^{6}}$ 

$\therefore x={{t}^{6}}\Rightarrow dx=6{{t}^{5}}$

$\therefore \int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=\int{\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}dx}=\int{\dfrac{6{{t}^{5}}}{{{t}^{2}}\left( 1+t \right)}dt}=6\int{\dfrac{{{t}^{3}}}{\left( 1+t \right)}dt}$ 

Now on dividing, we can obtain as shown below,

$\int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=6\int{\left\{ \left( {{t}^{2}}-t+1 \right)-\dfrac{1}{1+t} \right\}dt}$

$=6\int{\left[ \left( \dfrac{{{t}^{3}}}{3} \right)-\left( \dfrac{{{t}^{2}}}{2} \right)+t-\log \left| 1+t \right| \right]dt}$

$=2{{x}^{\dfrac{1}{2}}}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$

$=2\sqrt{x}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$, where $C$ is any arbitrary constant.


6. Integrate the expression: $\dfrac{\text{5x}}{\text{(x+1)}\left( {{\text{x}}^{\text{2}}}\text{+9} \right)}$.

Ans: The given expression is, $\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{Bx+C}{\left( {{x}^{2}}+9 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 5x=A\left( {{x}^{2}}+9 \right)+Bx+C\left( x+1 \right)$

$\Rightarrow 5x=A{{x}^{2}}+9A+B{{x}^{2}}+Bx+Cx+C$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+B=0\,\,\,...\left( 2 \right)$ 

$B+C=5\,\,\,...\left( 3 \right)$

$9A+C=0\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=\dfrac{-1}{2},\,B=\dfrac{1}{2},\,C=\dfrac{9}{2}$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}dx}=\int{\left\{ \dfrac{-1}{2\left( x+1 \right)}+\dfrac{x+9}{2\left( {{x}^{2}}+9 \right)} \right\}dx}$ 

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{2}\int{\dfrac{x}{{{x}^{2}}+9}dx}+\dfrac{9}{2}\int{\dfrac{1}{{{x}^{2}}+9}dx}$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\left( \dfrac{9}{2} \right)\left( \dfrac{1}{3} \right){{\tan }^{-1}}\left( \dfrac{x}{3} \right)$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\dfrac{3}{2}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)+C$, where $C$ is any arbitrary constant.


7. Integrate the expression, that is, $\dfrac{\text{sinx}}{\text{sin}\left( \text{x- }\!\!\alpha\!\!\text{ } \right)}$.

Ans: The given expression is, $\dfrac{\sin x}{\sin \left( x-\alpha  \right)}$. 

 Now, substitute $x-\alpha =t$ 

$\therefore dx=dt$

$\therefore \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=\int{\dfrac{\sin \left( t+\alpha  \right)}{\sin t}dt}=\int{\dfrac{\sin t\cos \alpha +\cos tsin\alpha }{\sin t}dt}=\int{\cos \alpha +\cot tsin\alpha dt}$$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=t\cos \alpha +\sin \alpha \log \left| \sin t \right|+{{C}_{1}}=\left( x-\alpha  \right)\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+{{C}_{1}}$ 

$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|-\alpha \cos \alpha +{{C}_{1}}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+C$

 where ${{C}_{1}},\,C$ are any arbitrary constants and $C={{C}_{1}}-\alpha \cos \alpha $.

 

8. Integrate the expression, that is, $\dfrac{{{\text{e}}^{\text{5logx}}}\text{-}{{\text{e}}^{\text{4logx}}}}{{{\text{e}}^{\text{3logx}}}\text{-}{{\text{e}}^{\text{2logx}}}}$.

Ans: The given expression is, $\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}$. 

 $\therefore \dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}=\dfrac{{{e}^{4\log x}}\left( {{e}^{\log x}}-1 \right)}{{{e}^{2\log x}}\left( {{e}^{\log x}}-1 \right)}={{e}^{2\log x}}={{x}^{2}}$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}dx}=\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C$, where $C$ is any arbitrary constant. 


9. Integrate the expression, that is, $\dfrac{\text{cosx}}{\sqrt{\text{4-si}{{\text{n}}^{\text{2}}}\text{x}}}$.

Ans: The given expression is, $\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}$. 

 Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

$\therefore \int{\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}dx}=\int{\dfrac{dt}{\sqrt{{{\left( 2 \right)}^{2}}-{{\left( t \right)}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{t}{2} \right)+C={{\sin }^{-1}}\left( \dfrac{\sin x}{2} \right)+C$, where $C$ is any arbitrary constant. 


10. Integrate the expression, that is, $\dfrac{\text{si}{{\text{n}}^{\text{8}}}\text{x-co}{{\text{s}}^{\text{8}}}\text{x}}{\text{1-2si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}}$.

Ans: The given expression is, $\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}$. 

 $\therefore \dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}$

$=\dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x\left( 1-{{\cos }^{2}}x \right)+{{\cos }^{2}}x\left( 1-{{\sin }^{2}}x \right)}$

$=\dfrac{-\left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{4}}x+{{\cos }^{4}}x}=-\cos 2x$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{-\cos 2xdx}=\dfrac{-\sin 2x}{2}+C$, where $C$ is any arbitrary constant.

 

11. Integrate the expression, that is, $\dfrac{\text{1}}{\text{cos}\left( \text{x+a} \right)\text{cos}\left( \text{x+b} \right)}$.

Ans: The given expression is, $\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}$. 

On multiplying and dividing by $\sin \left( \alpha -\beta  \right)$, the following can be obtained as shown below,

$\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( a-b \right)}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$ 

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left[ \left( x+a \right)-\left( x+b \right) \right]}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( x+a \right)\cos \left( x+b \right)-\cos \left( x+a \right)\sin \left( x+b \right)}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin (x+a)}{\cos (x+a)}-\dfrac{\sin \left( x+b \right)}{\cos (x+b)} \right]=\dfrac{1}{\sin (a-b)}\left( \tan (x+a)-\tan (x+b) \right)$

$\therefore \int{\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}dx}=\dfrac{1}{\sin (a-b)}\int{\left( \tan (x+a)-\tan (x+b) \right)dx}$ 

$\Rightarrow \int{\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}dx}=\dfrac{1}{\sin (a-b)}\left[ \log \left| \dfrac{\cos (x+b)}{\cos (x-a)} \right| \right]+C$, where $C$ is any arbitrary constant.


12. Integrate the expression, that is, $\dfrac{{{\text{x}}^{\text{3}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{8}}}}}$.

Ans: The given expression is, $\dfrac{{{x}^{3}}}{\sqrt{1-{{x}^{8}}}}$. 

 Now, substitute ${{x}^{4}}=t$ 

$\therefore 4{{x}^{3}}dx=dt$

$\therefore\int{\dfrac{{{x}^{3}}}{\sqrt{1-{{x}^{8}}}}dx}=\dfrac{1}{4}\int{\dfrac{dt}{\sqrt{1-{{\left( t \right)}^{2}}}}}=\dfrac{1}{4}{{\sin }^{-1}}t+C=\dfrac{1}{4}{{\sin }^{-1}}\left( {{x}^{4}} \right)+C$, where $C$ is any arbitrary constant. 


13. Integrate the expression, that is, $\dfrac{{{\text{e}}^{\text{x}}}}{\left( \text{1+}{{\text{e}}^{\text{x}}} \right)\left( \text{2+}{{\text{e}}^{\text{x}}} \right)}$.

Ans: The given expression is, $\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}$. 

 Now, substitute ${{e}^{x}}=t$ 

$\therefore {{e}^{x}}dx=dt$

$\therefore \int{\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}dx}=\int{\dfrac{dt}{\left( t+1 \right)\left( t+2 \right)}}=\int{\left[ \dfrac{1}{\left( t+1 \right)}-\dfrac{1}{\left( t+2 \right)} \right]dt}$

$\Rightarrow \int{\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}dx}=\log \left| t+1 \right|-\log \left| t+2 \right|+C=\log \left| \dfrac{{{e}^{x}}+1}{{{e}^{x}}+2} \right|+C$, where $C$ is any arbitrary constant. 


14. Integrate the expression, $\dfrac{\text{1}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)}\left( {{\text{x}}^{\text{2}}}\text{+4} \right)}$. 

Ans: The given expression is, $\dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}=\dfrac{Ax+B}{\left( {{x}^{2}}+1 \right)}+\dfrac{Cx+D}{\left( {{x}^{2}}+4 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 1=\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( {{x}^{2}}+9 \right)$

$\Rightarrow 1=A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+C{{x}^{3}}+Cx+D{{x}^{2}}+D$

On equating the coefficients of ${{x}^{3}},\,{{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=0\,\,\,...\left( 2 \right)$ 

$B+D=0\,\,\,...\left( 3 \right)$

$4A+C=0\,\,\,...\left( 4 \right)$

$4B+D=1\,\,\,...\left( 5 \right)$

And on solving these equations, the values of $A,\,B,\,C,D$ can be obtained as,

$A=0,\,B=\dfrac{1}{3},\,C=0,\,D=\dfrac{1}{3}$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}dx}=\dfrac{1}{3}\int{\left\{ \dfrac{1}{\left( {{x}^{2}}+1 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)}\, \right\}dx}$ 

$=\dfrac{1}{3}{{\tan }^{-1}}x-\dfrac{1}{\left( 3 \right)\left( 2 \right)}{{\tan }^{-1}}\dfrac{x}{2}+C$

$=\dfrac{1}{3}{{\tan }^{-1}}x-\dfrac{1}{6}{{\tan }^{-1}}\dfrac{x}{2}+C$, where $C$ is any arbitrary constant.


15. Integrate the expression, that is, $\text{co}{{\text{s}}^{\text{3}}}\text{x}{{\text{e}}^{\text{logsinx}}}$.

Ans: The given expression is, ${{\cos }^{3}}x{{e}^{\log \sin x}}$. 

Observe as shown below,

$\therefore {{\cos }^{3}}x{{e}^{\log \sin x}}={{\cos }^{3}}x\sin x$

Now, consider $cosx=t$ 

$\therefore -\sin xdx=dt$

$\therefore \int{{{\cos }^{3}}x{{e}^{\log \sin x}}dx}=\int{{{\cos }^{3}}x\sin xdx}=-\int{{{t}^{3}}dt}=\dfrac{-{{t}^{4}}}{4}+C=\dfrac{-{{\cos }^{4}}x}{4}+C$, where $C$ is any arbitrary constant. 


16. Integrate the expression, that is, ${{\text{e}}^{\text{3logx}}}{{\left( {{\text{x}}^{\text{4}}}\text{+1} \right)}^{\text{-1}}}$.

Ans: The given expression is, ${{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}$. 

Observe as shown below,

$\therefore {{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}=\dfrac{{{x}^{3}}}{\left( {{x}^{4}}+1 \right)}$

Now, consider ${{x}^{4}}+1=t$ 

$\therefore 4{{x}^{3}}dx=dt$

$\therefore \int{{{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}dx}=\int{\dfrac{{{x}^{3}}}{{{x}^{4}}+1}dx}=\dfrac{1}{4}\int{\dfrac{dt}{t}}=\dfrac{1}{4}\log \left| t \right|+C=\dfrac{\log \left| {{x}^{4}}+1 \right|}{4}+C$, where $C$ is any arbitrary constant. 


17. Integrate the expression, that is, $\text{f }\!\!'\!\!\text{ }\left( \text{ax+b} \right){{\left[ \text{f(ax+b)} \right]}^{\text{n}}}$.

Ans: The given expression is, $f'\left( ax+b \right){{\left[ f(ax+b) \right]}^{n}}$. 

Now, consider $\left[ f(ax+b) \right]=t$ 

$\therefore af'(ax+b)dx=dt$

$\therefore \int{f'\left( ax+b \right){{\left[ f(ax+b) \right]}^{n}}dx}=\dfrac{1}{a}\int{{{t}^{n}}dt}=\dfrac{{{t}^{n+1}}}{a\left( n+1 \right)}+C=\dfrac{{{\left[ f(ax+b) \right]}^{n+1}}}{a\left( n+1 \right)}+C$, where $C$ is any arbitrary constant. 


18. Integrate the expression, that is, $\dfrac{\text{1}}{\sqrt{\text{si}{{\text{n}}^{\text{3}}}\text{xsin}\left( \text{x+ }\!\!\alpha\!\!\text{ } \right)}}$.

Ans: The given expression is, $\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}$. 

$\therefore \dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}=\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin x\cos \alpha +\cos \alpha \sin x}}$

$=\dfrac{1}{\sqrt{{{\sin }^{4}}x\cos \alpha +{{\sin }^{3}}x\sin \alpha \cos x}}$

$=\dfrac{1}{{{\sin }^{2}}x\sqrt{\cos \alpha +\sin \alpha \cot x}}=\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cos \alpha +\sin \alpha \cot x}}$

Now, substitute $\cos \alpha +\sin \alpha \cot x=t$ 

$\therefore -\cos e{{c}^{2}}x\sin \alpha dx=dt$

$  \therefore \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}dx}=\int{\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cos \alpha +\sin \alpha \cot x}}}=\dfrac{-1}{\sin \alpha }\int{\dfrac{dt}{\sqrt{t}}}=\dfrac{-2\sqrt{t}}{\sin \alpha }+C $  

 $  \Rightarrow \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}dx}=\dfrac{-2\sqrt{\cos \alpha +\sin \alpha \cot x}}{\sin \alpha }+C=\dfrac{-2}{\sin \alpha }\sqrt{\cos \alpha +\dfrac{\sin \alpha \cos x}{\sin x}}+C $ 

, where $C$ is any arbitrary constant. 


19. Integrate the expression, that is, $\dfrac{\text{si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{-co}{{\text{s}}^{\text{-1}}}\sqrt{\text{x}}}{\text{si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{+co}{{\text{s}}^{\text{-1}}}\sqrt{\text{x}}}\text{,}\,\text{x}\in \left[ \text{0,}\,\text{1} \right]$.

Ans: The given expression is, $\dfrac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}},\,x\in \left[ 0,\,1 \right]$. 

 Assume, $I=\int{\dfrac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}dx}$ 

It is clearly known to us that, ${{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}=\dfrac{\pi }{2}$$\therefore I=\int{\dfrac{\dfrac{\pi }{2}-2{{\cos }^{-1}}\sqrt{x}}{\dfrac{\pi }{2}}dx}=\dfrac{2}{\pi }\int{\left( \dfrac{\pi }{2}-2{{\cos }^{-1}}\sqrt{x} \right)dx=}\dfrac{2}{\pi }.\dfrac{\pi }{2}\int{dx}-\dfrac{4}{\pi }\int{{{\cos }^{-1}}\sqrt{x}dx}$

$\Rightarrow I=x-\dfrac{4}{\pi }\int{{{\cos }^{-1}}\sqrt{x}dx}\,\,\,...\left( 1 \right)$

Assume, ${{I}_{1}}=\int{{{\cos }^{-1}}\sqrt{x}dx}$

Also, substitute $\sqrt{x}=t$ 

$\therefore dx=2tdt$

$\therefore {{I}_{1}}=\int{{{\cos }^{-1}}\sqrt{x}dx}=2\left[ {{\cos }^{-1}}t.\dfrac{{{t}^{2}}}{2}-\int{\left( \dfrac{-1}{\sqrt{1-{{t}^{2}}}} \right)\dfrac{{{t}^{2}}}{2}dt} \right]={{t}^{2}}{{\cos }^{-1}}t+\int{\left( \dfrac{{{t}^{2}}}{\sqrt{1-{{t}^{2}}}} \right)dt}$

$\Rightarrow {{I}_{1}}={{t}^{2}}{{\cos }^{-1}}t-\int{\sqrt{1-{{t}^{2}}}dt+}\int{\left( \dfrac{1}{\sqrt{1-{{t}^{2}}}} \right)dt}={{t}^{2}}{{\cos }^{-1}}t-\dfrac{t}{2}\sqrt{1-{{t}^{2}}}+\dfrac{\pi }{2}{{\sin }^{-1}}t$

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$I=x-\dfrac{4}{\pi }\left[ {{t}^{2}}{{\cos }^{-1}}t-\dfrac{t}{2}\sqrt{1-{{t}^{2}}}+\dfrac{\pi }{2}{{\sin }^{-1}}t \right]\,=x-\dfrac{4}{\pi }\left[ x{{\cos }^{-1}}\sqrt{x}-\dfrac{\sqrt{x}}{2}\sqrt{1-x}+\dfrac{\pi }{2}{{\sin }^{-1}}\sqrt{x} \right]$ 

$\Rightarrow I=x-\dfrac{4}{\pi }\left[ x\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\sqrt{x} \right)-\dfrac{1}{2}\sqrt{x-{{x}^{2}}}+\dfrac{\pi }{2}{{\sin }^{-1}}\sqrt{x} \right]$

$\Rightarrow I=-x+\dfrac{2}{\pi }\left[ \left( 2x-1 \right)\left( {{\sin }^{-1}}\sqrt{x} \right) \right]+\dfrac{2}{\pi }\sqrt{x-{{x}^{2}}}+C$

$\Rightarrow I=\dfrac{2\left( 2x-1 \right)}{\pi }{{\sin }^{-1}}\sqrt{x}+\dfrac{2}{\pi }\sqrt{x-{{x}^{2}}}+C$, where $C$ is any arbitrary constant.


20. Integrate the expression, that is, $\sqrt{\dfrac{\text{1-}\sqrt{\text{x}}}{\text{1+}\sqrt{\text{x}}}}$.

Ans: The given expression is, $\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}$. 

Assume, $I=\int{\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}dx}$ 

Now, substitute $x={{\cos }^{2}}\theta $ 

$\therefore dx=-2\sin \theta \cos \theta dt$

$\therefore I=\int{\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\left( -2\sin \theta \cos \theta  \right)d\theta }=-\int{\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\cos }^{2}}\dfrac{\theta }{2}}}\sin 2\theta d\theta }=-\int{\tan \dfrac{\theta }{2}2\sin \theta \cos \theta d\theta }$

$\Rightarrow I=-2\int{\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}\cos \theta d\theta }=-4\int{{{\sin }^{2}}\dfrac{\theta }{2}(2{{\cos }^{2}}\dfrac{\theta }{2}-1)d\theta }$

$\Rightarrow I=-8\int{{{\sin }^{2}}\dfrac{\theta }{2}{{\cos }^{2}}\dfrac{\theta }{2}d\theta }+4\int{{{\sin }^{2}}\dfrac{\theta }{2}d}\theta =2\int{{{\sin }^{2}}\dfrac{\theta }{2}d\theta }+4\int{{{\sin }^{2}}\dfrac{\theta }{2}d}\theta $

$\Rightarrow I=-2\int{\left( \dfrac{1-\cos 2\theta }{2} \right)d\theta }+4\int{\left( \dfrac{1-\cos \theta }{2} \right)d}\theta =-2\left[ \dfrac{\theta }{2}-\dfrac{\sin 2\theta }{2} \right]+4\left[ \dfrac{\theta }{2}-\dfrac{\sin \theta }{2} \right]+C$$\Rightarrow I=-\theta +\dfrac{\sin 2\theta }{2}+2\sin \theta +C=\theta +\sqrt{1-{{\cos }^{2}}\theta }\cos \theta -2\sqrt{1-{{\cos }^{2}}\theta }+C$

$\Rightarrow I={{\cos }^{-1}}\sqrt{x}+\sqrt{x(1-x)}-2\sqrt{1-x}+C=-2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x-{{x}^{2}}}+C$, where $C$ is any arbitrary constant. 


21. Integrate the expression, that is, $\dfrac{\text{2+sin2x}}{\text{1+cos2x}}{{\text{e}}^{\text{x}}}$.

Ans: The given expression is, $\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}$. 

Assume, $I=\int{\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}dx}$ 

$\Rightarrow I=\int{\dfrac{2+2\sin x\cos x}{2{{\cos }^{2}}x}{{e}^{x}}dx}=\int{\left( \dfrac{1+\sin x\cos x}{{{\cos }^{2}}x} \right){{e}^{x}}dx}=\int{\left( {{\sec }^{2}}x+\tan x \right){{e}^{x}}dx}$

Now, consider $f\left( x \right)=\tan x$ 

$\therefore f'\left( x \right)={{\sec }^{2}}xdx$

$\therefore I=\int{\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}}dx=\int{\left( f\left( x \right)+f'\left( x \right) \right){{e}^{x}}dx}={{e}^{x}}f\left( x \right)+C={{e}^{x}}\tan x+C$, where $C$ is any arbitrary constant. 


22. Integrate the expression: $\dfrac{{{\text{x}}^{\text{2}}}\text{+x+1}}{{{\text{(x+1)}}^{\text{2}}}\left( \text{x+2} \right)}$.

Ans: The given expression is, $\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{B}{\left( x+1 \right)}+\dfrac{C}{\left( x+2 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow {{x}^{2}}+x+1=A\left( x+1 \right)\left( x+2 \right)+B\left( x+2 \right)+C{{\left( x+1 \right)}^{2}}$

$\Rightarrow {{x}^{2}}+x+1=A\left( {{x}^{2}}+3x+2 \right)+B\left( x+2 \right)+C\left( {{x}^{2}}+2x+1 \right)$

$\Rightarrow {{x}^{2}}+x+1=\left( A+C \right){{x}^{2}}+x\left( 3A+B+2C \right)+\left( 2A+2B+C \right)$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=1\,\,\,...\left( 2 \right)$ 

$3A+B+2C=1\,\,\,...\left( 3 \right)$

$2A+2B+C=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-2,\,B=1,\,C=3$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}dx}=\int{\left\{ \dfrac{-2}{\left( x+1 \right)}+\dfrac{1}{\left( x+1 \right)}+\dfrac{3}{\left( x+2 \right)} \right\}dx}$ 

$=-2\int{\dfrac{1}{x+1}dx}+\int{\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx}+3\int{\dfrac{1}{x+2}dx}$

$=-2\log \left| x+1 \right|+3\log \left| x+2 \right|-\dfrac{1}{(x+1)}+C$, where $C$ is any arbitrary constant.


23. Integrate the expression, that is, $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\dfrac{\text{1-x}}{\text{1+x}}}$.

Ans: The given expression is, ${{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}$. 

Assume, $I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}dx}$ 

Now, consider $x=\cos \theta $ 

$\therefore dx=-\sin \theta d\theta $

$\therefore I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\left( -\sin \theta  \right)d\theta }=-\int{{{\tan }^{-1}}\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\cos }^{2}}\dfrac{\theta }{2}}}\sin \theta d\theta }=-\int{{{\tan }^{-1}}\tan \dfrac{\theta }{2}\sin \theta d\theta }$ $\Rightarrow I=-\dfrac{1}{2}\int{\theta \sin \theta d\theta }=-\dfrac{1}{2}\left[ \theta \left( -\cos \theta  \right)-\int{1.(-\cos \theta )d\theta } \right]=-\dfrac{1}{2}\left[ \theta \left( \cos \theta  \right)+\sin \theta  \right]$$\Rightarrow I=\dfrac{x}{2}{{\cos }^{-1}}x-\dfrac{1}{2}\sqrt{1-{{x}^{2}}}+C=\dfrac{1}{2}\left( x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}} \right)+C$, where $C$ is any arbitrary constant. 


24. Integrate the expression, that is, $\dfrac{\sqrt{{{\text{x}}^{\text{2}}}\text{+1}}\left[ \text{log}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{-2logx} \right]}{{{\text{x}}^{\text{4}}}}$.

Ans: The given expression is, $\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}$. 

Assume, $I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}$ $\Rightarrow I=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( {{x}^{2}}+1 \right)-\log {{x}^{2}} \right]=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}} \right) \right]=\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]$Consider $1+\dfrac{1}{{{x}^{2}}}=t$ 

$\therefore \dfrac{-2}{{{x}^{3}}}dx=dt$

Now, integrate the given expression as shown below $\therefore I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}=\int{\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]dx}=\dfrac{-1}{2}\int{{{t}^{\dfrac{1}{2}}}\log tdt}+C$Using integration by parts, it can be obtained that,

$I=\dfrac{-1}{2}\left[ \log t.\int{{{t}^{\dfrac{1}{2}}}dt-\left\{ \left( \dfrac{d}{dt}\log t \right)\int{{{t}^{\dfrac{1}{2}}}dt} \right\}dt} \right]=\dfrac{-1}{2}\left[ \log t.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\int{\dfrac{1}{t}.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}dt} \right]$ 

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ 1+\dfrac{1}{{{x}^{2}}} \right]\left( \log \left( 1+\dfrac{1}{{{x}^{2}}} \right)-\dfrac{2}{3} \right)+C$, where $C$ is any arbitrary constant. 


25. Find the value of the expression, that is, $\int_{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}^{\text{ }\!\!\pi\!\!\text{ }}{{{\text{e}}^{\text{x}}}\left( \dfrac{\text{1-sinx}}{\text{1-cosx}} \right)\text{dx}}$.

Ans: The given expression is, $\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$. 

Assume, $I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$ 

$\Rightarrow I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{+2{{\sin }^{2}}\dfrac{x}{2}} \right)dx=}\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{\cos e{{c}^{2}}\dfrac{x}{2}}{2}-\cot \dfrac{x}{2} \right)dx}$

Now, substitute $f\left( x \right)=-\cot \dfrac{x}{2}$ 

$\therefore f'\left( x \right)=-\left( -\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2} \right)dx=\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2}dx$

$\therefore I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( f(x)+f'(x) \right)dx}=\left[ {{e}^{x}}f(x) \right]_{\dfrac{\pi }{2}}^{\pi }=\left[ {{e}^{x}}\cot \dfrac{x}{2} \right]_{\dfrac{\pi }{2}}^{\pi }$

$\Rightarrow I=\left[ {{e}^{\pi }}\cot \dfrac{\pi }{2}-{{e}^{\dfrac{\pi }{2}}}\cot \dfrac{\pi }{4} \right]=\left[ 0-{{e}^{\dfrac{\pi }{2}}} \right]=-{{e}^{\dfrac{\pi }{2}}}$


26. Find the value of the expression, that is, $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\dfrac{\text{sinxcosx}}{\text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\dfrac{\sin x\cos x}{{{\cos }^{4}}x}}{\dfrac{{{\sin }^{4}}x+{{\cos }^{4}}x}{{{\cos }^{4}}x}}dx}=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}dx}$

Now, substitute ${{\tan }^{2}}x=t$ 

$\therefore 2\tan x{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{4},\,t=1$.

$\therefore I=\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{1+{{t}^{2}}}}=\dfrac{1}{2}\left[ {{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ {{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{1}{2}\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}$


27. Find the value of the expression, that is, $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{x}}{\text{co}{{\text{s}}^{\text{2}}}\text{x+4si}{{\text{n}}^{\text{2}}}\text{x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4\left( 1-{{\cos }^{2}}x \right)}dx}=\dfrac{-1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-4-3{{\cos }^{2}}x}{-3{{\cos }^{2}}x+4}dx}$

$\Rightarrow I=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-3{{\cos }^{2}}x}{4-3{{\cos }^{2}}x}dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4}{4-3{{\cos }^{2}}x}dx}=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4{{\sec }^{2}}x-3}dx}$

$\Rightarrow I=\dfrac{-1}{3}\left[ x \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4\left( 1+{{\tan }^{2}}x \right)-3}dx}=\dfrac{-\pi }{6}+\dfrac{2}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}\,\,\,...\left( 1 \right)$

Observe, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}$

Now, substitute $2\tan x=t$ 

$\therefore 2{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=\infty $. 

$\therefore \int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}=\int_{0}^{\infty }{\dfrac{dt}{\left( 1+{{t}^{2}} \right)}dx}=\left[ {{\tan }^{-1}}\left( t \right) \right]_{0}^{\infty }=\left[ {{\tan }^{-1}}\left( \infty  \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{\pi }{2}$Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=-\dfrac{\pi }{6}+\dfrac{2}{3}\left( \dfrac{\pi }{2} \right)=-\dfrac{\pi }{6}+\dfrac{2\pi }{6}=\dfrac{\pi }{6}$


28. Find the value of the expression, that is, $\int_{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\dfrac{\text{sinx+cosx}}{\sqrt{\text{sin2x}}}\text{dx}}$.

Ans: The given expression is, $\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$. 

Assume, $I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{-\left( -1+1-2\sin x\cos x \right)}}dx}=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)}}dx}$

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}$

Now, substitute $\left( \sin x-\cos x \right)=t$ 

$\therefore (\sin x+\cos x)dx=dt$

And also when $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 

$\therefore I=\int_{\dfrac{1-\sqrt{3}}{2}}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\int_{-\left( \dfrac{-1+\sqrt{3}}{2} \right)}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}$

As $\dfrac{1}{\sqrt{1-{{\left( -t \right)}^{2}}}}=\dfrac{1}{\sqrt{1-{{t}^{2}}}}$, it can be thus obtained that $\dfrac{1}{\sqrt{1-{{t}^{2}}}}$ is an even function, 

$\therefore \int_{-a}^{a}{f\left( x \right)dx}=2\int_{0}^{a}{f\left( x \right)dx}$ 

$\therefore I=2\int_{0}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\left[ 2{{\sin }^{-1}}t \right]_{0}^{\dfrac{\sqrt{3}-1}{2}}=2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-1}{2} \right)$ $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 


29. Find the value of the expression, that is, $\int_{\text{0}}^{\text{1}}{\dfrac{\text{dx}}{\sqrt{\text{1+x}}\text{-}\sqrt{\text{x}}}}$.

Ans: The given expression is, $\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$. 

Assume, $I=\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$ 

$\Rightarrow I=\int_{0}^{1}{\dfrac{1}{\sqrt{1+x}-\sqrt{x}}\times \dfrac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}dx}=\int_{0}^{1}{\dfrac{\sqrt{1+x}+\sqrt{x}}{1+x-x}dx}$

$\Rightarrow I=\int_{0}^{1}{\sqrt{1+x}dx}+\int_{0}^{1}{\sqrt{x}dx}=\dfrac{2}{3}\left[ {{\left( 1+x \right)}^{\dfrac{2}{3}}} \right]_{0}^{1}+\dfrac{2}{3}\left[ {{\left( x \right)}^{\dfrac{3}{2}}} \right]_{0}^{1}=\dfrac{2}{3}\left[ {{\left( 2 \right)}^{\dfrac{2}{3}}}-1 \right]+\dfrac{2}{3}=\dfrac{4\sqrt{2}}{3}$


30. Find the value of the expression, that is, $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{}}{\text{4}}}{\dfrac{\text{sinx+cosx}}{\text{9+16sin2x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$ 

Now, substitute $\sin x-\cos x=t$ 

$\therefore \left( \cos x+\sin x \right)dx=dt$

And also when $x=0,\,t=-1$ and when $x=\dfrac{\pi }{4},\,t=0$.

$\therefore {{\left( \sin x-\cos x \right)}^{2}}={{t}^{2}}$

$\Rightarrow 1-2\sin x\cos x={{t}^{2}}$

$\Rightarrow 1-\sin 2x={{t}^{2}}$

$\Rightarrow \sin 2x=1-{{t}^{2}}$

$\therefore I=\int_{-1}^{0}{\dfrac{dt}{9+16\left( 1-{{t}^{2}} \right)}}=\int_{-1}^{0}{\dfrac{dt}{25-16{{t}^{2}}}}=\int_{-1}^{0}{\dfrac{dt}{{{\left( 5 \right)}^{2}}-{{\left( 4t \right)}^{2}}}}$

$\Rightarrow I=\dfrac{1}{4}\left[ \dfrac{1}{2\left( 5 \right)}\log \left| \dfrac{5+4t}{5-4t} \right| \right]_{-1}^{0}=\dfrac{1}{40}\left[ \log \left| 1 \right|-\log \left| \dfrac{1}{9} \right| \right]=\dfrac{1}{40}\log \left| 9 \right|$


31. Find the value of the expression, that is, $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{sin2xta}{{\text{n}}^{\text{-1}}}\left( \text{sinx} \right)\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}=\int_{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=1$. 

$\therefore I=2\int_{0}^{1}{t{{\tan }^{-1}}\left( t \right)dt}$

Observe, $\int{t{{\tan }^{-1}}\left( t \right)dt}$

$\therefore \int{t{{\tan }^{-1}}\left( t \right)dt}={{\tan }^{-1}}t\int{tdt}-\int{\left\{ \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int{tdt} \right\}dt}={{\tan }^{-1}}t.\dfrac{{{t}^{2}}}{2}-\int{\dfrac{1}{1+{{t}^{2}}}.\dfrac{{{t}^{2}}}{2}dt}$

$=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{1dt}+\int{\dfrac{1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t$

$\therefore \int_{0}^{1}{t.{{\tan }^{-1}}tdt}=\left[ \dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-1+\dfrac{\pi }{4} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}$

Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=2\left[ \dfrac{\pi }{2}-\dfrac{1}{2} \right]=\dfrac{\pi }{2}-1$


32. Find the value of the expression, that is, $\int_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\dfrac{\text{xtanx}}{\text{secx+tanx}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\pi }{\dfrac{x\tan x}{\sec x+\tan x}dx}\,\,\,...\left( 1 \right)$. 

Assume, $I=\int_{0}^{\pi }{\dfrac{x\tan x}{\sec x+\tan x}dx}$ 

$\Rightarrow I=\int_{0}^{\pi }{\dfrac{\left( \pi -x \right)\tan \left( \pi -x \right)}{\sec \left( \pi -x \right)+\tan \left( \pi -x \right)}dx\,\,\,\,\,\,\left[ \because \int_{0}^{a}{f\left( x \right)dx}=\int_{0}^{a}{f\left( a-x \right)dx} \right]}$

$\Rightarrow I=\int_{0}^{\pi }{\dfrac{-\left( \pi -x \right)\tan \left( x \right)}{-(\sec \left( x \right)+\tan \left( x \right))}dx}=\int_{0}^{\pi }{\dfrac{\left( \pi -x \right)\tan \left( x \right)}{\sec \left( x \right)+\tan \left( x \right)}dx}\,\,\,...\left( 2 \right)\,$

Now, on adding equations $\left( 1 \right)$ and $\left( 2 \right)$, it can be obtained that, 

$2I=\int_{0}^{\pi }{\dfrac{x\tan x}{\sec x+\tan x}dx}=\pi \int_{0}^{\pi }{\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}}dx}=\pi \int_{0}^{\pi }{\dfrac{\sin x}{1+\sin x}dx}$

$\Rightarrow 2I=\pi \int_{0}^{\pi }{\dfrac{\sin x+1-1}{1+\sin x}dx}=\pi \int_{0}^{\pi }{dx}-\pi \int_{0}^{\pi }{\dfrac{1}{1+\sin x}dx}=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\dfrac{1-\sin x}{{{\cos }^{2}}x}dx}$

$\Rightarrow 2I=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}={{\pi }^{2}}-\pi \left[ \tan x-\sec x \right]_{0}^{\pi }$

$\Rightarrow 2I={{\pi }^{2}}-\pi \left[ 0-\left( -1 \right)-0+1 \right]={{\pi }^{2}}-2\pi $

$\Rightarrow I=\dfrac{\pi \left( \pi -2 \right)}{2}$


33. Find the value of the expression, that is, $\int_{\text{1}}^{\text{4}}{\left[ \left| \text{x-1} \right|\text{+}\left| \text{x-2} \right|\text{+}\left| \text{x-3} \right| \right]\text{dx}}$.

Ans: The given expression is, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$. 

Assume, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$ 

$\Rightarrow I=\int_{1}^{4}{\left| x-1 \right|dx}+\int_{1}^{4}{\left| x-2 \right|dx}+\int_{1}^{4}{\left| x-3 \right|dx}$

$\therefore I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\,\,\,...\left( 1 \right)$

where, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx},\,{{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx},\,{{I}_{3}}=+\int_{1}^{4}{\left| x-3 \right|dx}$ 

Now, consider, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx}$, where $\left( x-1 \right)\ge 0\,\forall \,1\le x\le 4$  $\therefore {{I}_{1}}=\int_{1}^{4}{\left( x-1 \right)dx}=\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{1}^{4}=\left[ 8-4-\dfrac{1}{2}+1 \right]=\dfrac{9}{2}\,\,\,...\left( 2 \right)\,$

Again, consider, ${{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx}$, where $\left( x-2 \right)\ge 0\,\forall \,2\le x\le 4$ and $\left( x-2 \right)\le 0\,\forall \,1\le x\le 2$.

 $\therefore {{I}_{2}}=\int_{1}^{2}{\left( 2-x \right)dx}+\int_{2}^{4}{\left( x-2 \right)dx}=\left[ 2x-\dfrac{{{x}^{2}}}{2} \right]_{1}^{4}+\left[ \dfrac{{{x}^{2}}}{2}-2x \right]_{2}^{4}$

$\Rightarrow {{I}_{2}}=\left[ 4-2-2+\dfrac{1}{2} \right]+\left[ 8-8-2+4 \right]=\dfrac{1}{2}+2=\dfrac{5}{2}\,\,\,...\left( 3 \right)\,$

Also, consider, ${{I}_{3}}=\int_{1}^{4}{\left| x-3 \right|dx}$, where $\left( x-3 \right)\ge 0\,\forall \,3\le x\le 4$ and $\left( x-3 \right)\le 0\,\forall \,1\le x\le 3$.

 $\therefore {{I}_{3}}=\int_{1}^{3}{\left( 3-x \right)dx}+\int_{3}^{4}{\left( x-3 \right)dx}=\left[ 3-\dfrac{{{x}^{2}}}{2} \right]_{1}^{3}+\left[ \dfrac{{{x}^{2}}}{2}-3x \right]_{3}^{4}$

$\Rightarrow {{I}_{3}}=\left[ 9-\dfrac{9}{2}-3+\dfrac{1}{2} \right]+\left[ 8-12-\dfrac{9}{2}+9 \right]=2+\dfrac{1}{2}=\dfrac{5}{2}\,\,\,...\left( 4 \right)\,$

Now, from equations $\left( 1 \right)$, $\left( 2 \right)$, $\left( 3 \right)$ and $\left( 4 \right)$ it can be obtained that, 

$I=\dfrac{9}{2}+\dfrac{5}{2}+\dfrac{5}{2}=\dfrac{19}{2}$

$\Rightarrow 2I=\pi \int_{0}^{\pi }{\dfrac{\sin x+1-1}{1+\sin x}dx}=\pi \int_{0}^{\pi }{dx}-\pi \int_{0}^{\pi }{\dfrac{1}{1+\sin x}dx}=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\dfrac{1-\sin x}{{{\cos }^{2}}x}dx}$

$\Rightarrow 2I=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}={{\pi }^{2}}-\pi \left[ \tan x-\sec x \right]_{0}^{\pi }$

$\Rightarrow 2I={{\pi }^{2}}-\pi \left[ 0-\left( -1 \right)-0+1 \right]={{\pi }^{2}}-2\pi $

$\Rightarrow I=\dfrac{\pi \left( \pi -2 \right)}{2}$


34. Prove the following equation: $\int_{\text{1}}^{\text{3}}{\dfrac{\text{dx}}{{{\text{x}}^{\text{2}}}\left( \text{x+1} \right)}}\text{=}\dfrac{\text{2}}{\text{3}}\text{+log}\dfrac{\text{2}}{\text{3}}$.

Ans: The given equation is, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\dfrac{2}{3}+\log \dfrac{2}{3}$. 

Assume, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}$ 

Now consider it as shown below,

$\therefore \dfrac{1}{{{x}^{2}}\left( x+1 \right)}=\dfrac{A}{x}+\dfrac{B}{{{x}^{2}}}+\dfrac{C}{\left( x+1 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 1=Ax\left( x+1 \right)+B\left( x+1 \right)+C\left( {{x}^{2}} \right)$

$\Rightarrow 1=A{{x}^{2}}+Ax+Bx+B+C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=0\,\,\,...\left( 2 \right)$ 

$A+B=0\,\,\,...\left( 3 \right)$

$B=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-1,\,B=1,\,C=1$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$I=\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\int_{1}^{3}{\left\{ \dfrac{-1}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{\left( x+1 \right)} \right\}dx}$ 

$\Rightarrow I=\left[ -\log x-\dfrac{1}{x}+\log \left( x+1 \right) \right]_{1}^{3}=\left[ \log \left( \dfrac{x+1}{x} \right)-\dfrac{1}{x} \right]_{1}^{3}=\log \left( \dfrac{4}{3} \right)-\dfrac{1}{3}-\log \left( 2 \right)+1$

$\Rightarrow I=\log 4-\log 3-\log 2+\dfrac{2}{3}=\log 2-\log 3+\dfrac{2}{3}=\log \left( \dfrac{2}{3} \right)+\dfrac{2}{3}$

Henceforth, it can be clearly proved.


35. Prove the following equation: $\int_{\text{0}}^{\text{4}}{\text{x}{{\text{e}}^{\text{x}}}\text{dx}}\text{=1}.

Ans: The given equation is, $\int_{0}^{4}{x{{e}^{x}}dx}=1$. 

Assume, $I=\int_{0}^{4}{x{{e}^{x}}dx}$ 

Using integration by parts, it can be obtained that,

$I=x\int_{0}^{4}{{{e}^{x}}dx}-\int_{0}^{4}{\left\{ \left( \dfrac{d\left( x \right)}{dx} \right)\int{{{e}^{x}}} \right\}}=\left[ x{{e}^{x}} \right]_{0}^{1}-\left[ {{e}^{x}} \right]_{0}^{1}=e-e+1=1$ 

Henceforth, it can be clearly proved.


36. Prove the following equation: $\int_{\text{1}}^{\text{-1}}{{{\text{x}}^{\text{17}}}\text{co}{{\text{s}}^{\text{4}}}\text{xdx}}\text{=0}$.

Ans: The given equation is, $\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$. 

Assume, $I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}$ 

Now, consider $f(x)={{x}^{17}}{{\cos }^{4}}x$ 

$\therefore f\left( -x \right)={{\left( -x \right)}^{17}}{{\cos }^{4}}\left( -x \right)=-{{x}^{17}}{{\cos }^{4}}x=-f\left( x \right)$

$\Rightarrow f\left( x \right)$ is an odd function and henceforth it is clearly known to us that when $f(x)$ is an odd function, then $\int_{-a}^{a}{f\left( x \right)dx}=0$. 

$\therefore I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$

Henceforth, it can be clearly proved.


37. Prove the following equation: $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{3}}}\text{xdx}}\text{=}\dfrac{\text{2}}{\text{3}}$.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}=\dfrac{2}{3}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\left( 1-{{\cos }^{2}}x \right)\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\sin xdx}-\int_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}x\sin xdx}$

$\Rightarrow I=\left[ -\cos x \right]_{0}^{\dfrac{\pi }{2}}+\left[ \dfrac{co{{s}^{3}}x}{3} \right]_{0}^{\dfrac{\pi }{2}}=1-\dfrac{1}{3}=\dfrac{2}{3}$

Henceforth, it can be clearly proved.


38. Prove the following equation: $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{2ta}{{\text{n}}^{\text{3}}}\text{xdx}}\text{=1-log2}$.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}=1-\log 2$. 

Assume, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{2}}x\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\left( 1-{{\sec }^{2}}x \right)\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\tan xdx}-\int_{0}^{\dfrac{\pi }{4}}{{{\sec }^{2}}x\tan xdx}$

$\Rightarrow I=2\left[ \dfrac{{{\tan }^{2}}x}{2} \right]_{0}^{\dfrac{\pi }{4}}+2\left[ \log \cos x \right]_{0}^{\dfrac{\pi }{4}}=1+2\left[ \log \cos \dfrac{\pi }{4}-\log \cos 0 \right]_{0}^{\dfrac{\pi }{2}}=1-\log 2-\log 1$

$\Rightarrow I=1-\log 2$

Henceforth, it can be clearly proved.


39. Prove the following equation: $\int_{\text{0}}^{\text{1}}{\text{si}{{\text{n}}^{\text{-1}}}\text{xdx}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-1}$.

Ans: The given equation is, $\int_{0}^{1}{{{\sin }^{-1}}xdx}=\dfrac{\pi }{2}-1$. 

Assume, $I=\int_{0}^{1}{{{\sin }^{-1}}xdx}$ 

$\Rightarrow I=\int_{0}^{1}{{{\sin }^{-1}}x.1dx}$

Using integration by parts, it can be obtained that,

$I=\left[ {{\sin }^{-1}}x.x \right]_{0}^{1}-\int_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}xdx}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{\left( -2x \right)}{\sqrt{1-{{x}^{2}}}}dx}$ 

Now, substitute $1-{{x}^{2}}=t$ 

$\therefore \left( -2x \right)dx=dt$

And also when $x=0,\,t=1$ and when $x=1,\,t=0$.

$\therefore I=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{\sqrt{t}}}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\left[ 2\sqrt{t} \right]_{1}^{0}={{\sin }^{-1}}1-\sqrt{1}=\dfrac{\pi }{2}-1$ Henceforth, it can be clearly proved.


40. Evaluate $\int_{\text{0}}^{\text{1}}{{{\text{e}}^{\text{2-3x}}}}$ as a limit of sum.

Ans: Assume, $I=\int_{0}^{1}{{{e}^{2-3x}}}$ 

It is clearly known that $\int_{a}^{b}{f(x)dx}=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{a}\left[ f\left( a \right)+f(a+h)+...+f(a+(n-1)h) \right],\,$where, $h=\dfrac{b-a}{n}$ 

In this case, $a=0,\,b=1,\,f\left( x \right)={{e}^{2-3x}}$ 

$\Rightarrow h=\dfrac{1-0}{n}=\dfrac{1}{n}$  

$\therefore \int_{0}^{1}{{{e}^{2-3x}}dx}=\left( 1-0 \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f\left( 0 \right)+f(0+h)+...+f(0+(n-1)h) \right]$

$\Rightarrow \int_{0}^{1}{{{e}^{2-3x}}dx}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{2}}+{{e}^{2-3h}}+...+{{e}^{2-3\left( n-1 \right)h}} \right]=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{2}}\left( 1+{{e}^{-3h}}+{{e}^{-6h}}...+{{e}^{-3\left( n-1 \right)h}} \right) \right]$

$\Rightarrow \int_{0}^{1}{{{e}^{2-3x}}dx}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{2}}\left( \dfrac{1-{{e}^{\dfrac{-3}{n}.n}}}{1-{{e}^{\dfrac{-3}{n}}}} \right) \right]=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{2}}\left( \dfrac{1-{{e}^{-3}}}{1-{{e}^{\dfrac{-3}{n}}}} \right) \right]={{e}^{2}}\left( {{e}^{-3}}-1 \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \dfrac{1}{{{e}^{\dfrac{-3}{n}}}-1} \right]$

$\Rightarrow \int_{0}^{1}{{{e}^{2-3x}}dx}={{e}^{2}}\left( {{e}^{-3}}-1 \right)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{-1}{3}\left[ \dfrac{\dfrac{-3}{n}}{{{e}^{\dfrac{-3}{n}}}-1} \right]=\dfrac{-{{e}^{2}}\left( {{e}^{-3}}-1 \right)}{3}=\dfrac{-{{e}^{-1}}+{{e}^{2}}}{3}=\dfrac{1}{3}\left( {{e}^{2}}-\dfrac{1}{e} \right)$


41. The expression, that is, $\int{\dfrac{\text{dx}}{{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}}}$ is equal to

A. $\text{ta}{{\text{n}}^{\text{-1}}}\left( {{\text{e}}^{\text{x}}} \right)\text{+C}$

B. $\text{ta}{{\text{n}}^{\text{-1}}}\left( {{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

C. $\text{log}\left( {{\text{e}}^{\text{x}}}\text{-}{{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

D. $\text{log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

Ans: The given expression is, $\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$. 

Assume, $I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$ 

Now, consider ${{e}^{x}}=t$ 

$\therefore {{e}^{x}}dx=dt$

$\therefore I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}=\int{\dfrac{1}{1+{{t}^{2}}}dt}=\int{{{\tan }^{-1}}\operatorname{t}dt}+C$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (A). 


42. The expression, that is, $\int{\dfrac{\text{cos2x}}{{{\left( \text{sinx+cosx} \right)}^{\text{2}}}}\text{dx}}$ is

A. $\dfrac{\text{-1}}{\text{sinx+cosx}}\text{+C}$

B. $\text{log}\left| \text{sinx+cosx} \right|\text{+C}$ 

C. $\text{log}\left| \text{sinx-cosx} \right|\text{+C}$ 

D. $\dfrac{\text{1}}{{{\left( \text{sinx+cosx} \right)}^{\text{2}}}}\text{+C}$ 

Ans: The given expression is, $\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$. 

Assume, $I=\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$ 

$\Rightarrow I=\int{\dfrac{(\sin x+\cos x)(\cos x-\sin x)}{{{\left( \sin x+\cos x \right)}^{2}}}dx}=\int{\dfrac{(\cos x-\sin x)}{\left( \sin x+\cos x \right)}dx}$

Now, substitute $\left( \sin x+\cos x \right)=t$ 

$\therefore (\cos x-\sin x)dx=dt$

$\therefore I=I=\int{\dfrac{1}{t}dt=\log \left| t \right|+C=\log \left| \cos x+\sin x \right|+C}$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (B). 


43. If $\text{f}\left( \text{a+b-x} \right)\text{=f}\left( \text{x} \right)\text{,}\,$then $\int_{\text{a}}^{\text{b}}{\text{xf}\left( \text{x} \right)\text{dx}}$ is equal to

A. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{b-x} \right)\text{dx}}$

B. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{b+x} \right)\text{dx}}$ 

C. $\dfrac{\text{b-a}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{x} \right)\text{dx}}$ 

D. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{x} \right)\text{dx}}$ 

Ans: Assume, $I=\int_{a}^{b}{xf\left( x \right)dx}\,\,\,...\left( 1 \right)$ 

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( a+b-x \right)dx}\,\,\,\left[ \because \int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx} \right]$

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( x \right)dx}=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}-I$   using $\left( 1 \right)$

$\Rightarrow 2I=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}$

$\Rightarrow I=\dfrac{\left( a+b \right)}{2}\int_{a}^{b}{f\left( x \right)dx}$

Hence, the correct answer is option (D). 


44. The value of $\int_{\text{0}}^{\text{1}}{\text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{2x-1}}{\text{1+x-}{{\text{x}}^{\text{2}}}} \right)\text{dx}}$ is equal to

A. $\text{1}$

B. $\text{0}$ 

C. $\text{-1}$ 

D. $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ 

Ans: Assume, $I=\int_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{2x-1}{1+x-{{x}^{2}}} \right)dx}$ $\Rightarrow I=\int_{0}^{1}{{{\tan }^{-1}}\left( \dfrac{x-\left( 1-x \right)}{1+x\left( 1-x \right)} \right)dx}=\int_{0}^{1}{{{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( 1-x \right)dx}\,\,\,...\left( 1 \right)\,$

$\Rightarrow I=0$

Hence, the correct answer is option (D). 

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

7.1 Introduction

In Chapter 7 Maths Class 12, the introduction part delves into the history of differential calculus and how integral calculus came into existence. You would learn how the original idea of derivative came about to solve the problem of defining tangent lines to the graphs of functions. It also deals with calculating the slope of these tangent lines.

 

This chapter will introduce the prime concept of the integral calculus, which involves trying to find the area bounded by the graph of these functions. You would learn how integration and differentiation are interrelated, and integration is the reverse process of differentiation. In differentiation, we are given a function, and we need to figure out the differentiation of that function. Conversely, in integration, you are given the differential of a function, and we need to find the function.

 

Class 12 Maths Chapter 7 – Integrals 

What are Integrals?

Integrals are defined as the values of functions that are determined by the process of integration. The process of calculating f(x) from f'(x) is called integration.

 

What are the Topics Covered in Class 12 Maths Chapter 7?

Here is the list of different topics covered in the exercises of NCERT Class 12 Maths Chapter 7.

 

Topics Covered in Exercise 7.1

  • Introduction to Factorization

  • Integral as an inverse process of differentiation

  • Geometric interpretation of Indefinite integral

  • Properties of Indefinite Integral

  • Comparison Between Differentiation and Integration

 

Topics Covered in Exercise 7.2

  • Method of Integration: Integration By Substitution, Integration Using Partial Fractions, and Integration By Parts

 

Topics Covered in Exercise 7.3

  • Integration Using Trigonometric Identities

 

Topics Covered in Exercise 7.4

  • Integration of some particular function

 

Topics Covered in Exercise 7.5

  • Integration by partial function

 

Topics Covered in Exercise 7.6

  • Integration By Parts

 

Topics Covered in Exercise 7.7

  • Integration of Some More Types

 

Topics Covered in Exercise 7.8

  • Definite Integral

  • Definite Integral as the limit of Sum

 

Topics Covered in Exercise 7.9

  • First and Second Fundamental Theorem of Integral Calculus

 

Topics Covered in Exercise 7.10

  • Evaluation of Definite Integral By Substitution

 

Topics Covered in Exercise 7.11

  • Properties of Definite Integral 

 

Other Key Concepts Discussed in This Chapter are

  • What are indefinite integrals and how they are denoted 

  • What are definite integrals

  • How the connection between the fundamentals of indefinite integrals and definite integrals serve as a practical tool for engineering and science. This connection is termed as the fundamental theorem of calculus.

With many different kinds of exercises and solved problems in this NCERT Solutions Class 12 Maths Chapter 7 you can easily revise all that you learn on this topic.

 

7.2 Integration as a Reverse Process of Differentiation

In NCERT Solutions for Class 12 Chapter 7 Applications of Derivatives, you would learn how one can find the integral of a function given its derivative. The other name of integral is also anti-differentiation. You would understand terms like “arbitrary constant” which are varied to get different integrals of any given function. This “arbitrary constant” is also known as “constant of integration” and for any arbitrary real number K, ∫f(x) dx = F(x) + K. This is an indefinite integral. You will get acquainted with a lot of basic formulas and properties of the indefinite integral in this section and how it is interpreted geometrically.

 

This chapter would take you through a comparison between integration and differentiation and you will realize a few basics about these 2 parts of calculus like:

  • Differentiation and integration are operations on functions

  • They both satisfy linearity properties.

  • There could be functions that are neither differentiable nor integrable.

  • They differ in the sense that differentiation of a function (wherever it exists) is a unique function while integral is not.

  • Exercise 7.1 Solutions: 22 Questions (21 Short Questions, 1 MCQ)

 

7.3 Methods of Integration

In this part of Integration NCERT Solutions, you will gain knowledge on other methods of finding integral apart from the method of inspection discussed in the above part. You would realize how this method is useful when an inspection is not suitable for many different kinds of functions. Here you will understand 3 different ways of integration i.e.:

  • By Substitutions – This is used when a variable is substituted by another suitable variable it makes the integration process easier. So, if I = ∫f(x)dx and we put x = g(t) then we can write I = ∫ f

  • g(t)

  • g(t) g'(t) dt

  • Using Partial Functions – If there are 2 polynomials p1(x) and p2(x) on x where p2(x) <>0 and degrees of p1(x) > p2(x) then we can say:

p1(x)/ p2(x) = t(x) + p’(x)/ p2(x). Here t(x) is a polynomial in x which is easily integrated and degree of p’(x) < degree of p2(x). 

  • By Parts – If there are 2 functions f1(x) and f2(x) then we can say ∫

  • f1(x)f2(x)

  • f1(x)f2(x) dx = f1(x) ∫f2(x)] dx – ∫{f'(x) ∫f2(x)] dx} dx. This method is not applicable to the product of functions in all the cases like it will not work for ∫√x sin x dx.

You would further learn how to integrate using trigonometric identities for trigonometric functions with the use of many known trigonometric identities like:

  • Sin2 Ө + Cos2 Ө = 1

  • 2 Sin Ө Cos Ө = Sin 2 Ө

  • 2 Sin Ө1. Cos Ө2 = Sin (Ө1 + Ө2) + Sin (Ө1 - Ө2)

  • Exercise 7.2 Solutions: 39 Questions (37 Short Questions, 2 MCQs)

  • Exercise 7.3 Solutions: 24 Questions (22 Short Questions, 2 MCQs)

 

7.4 Integrals of Some Particular Functions

In this part of NCERT Class 12 Maths Chapter 7 Solutions, you would learn many important integral formulas and how to apply them for many other standard integrals that are related to them. These short cut techniques provided in our solutions to problems on Integration Class 12 will help immensely in tackling many types of questions really fast.

 

You will also understand how to prove these standard integrals, some of those are mentioned below:

  • ∫ dx/ (x2 – b2) = 1/2b log |x – b/x + a| + C

  • ∫ dx/ (x2 + b2) = 1/b tan-1 x/b + C

  • ∫ dx/ (√x2 + b2) = log |x + √x2 + b2| + C

  • Exercise 7.4 Solutions: 25 Questions (23 Short Questions, 2 MCQs)

 

7.5 Integration by Partial Functions

This unit of Chapter 7 Class 12 Maths NCERT solutions further expands what you learned in unit 7.3 about integration by partial functions. In this section, you would learn how an improper rational function can be converted into a proper rational function by a long division process. An improper rational function is the one where for 2 polynomials p1(x) and p2(x) on x,  p2(x) <>0 and degrees of p1(x) > p2(x) t.

 

You will learn here how to disintegrate an equation into parts and find the integral of each of the parts. This is known as partial fraction decomposition. 

  • Exercise 7.5 Solutions: 23 Questions (21 Short Questions, 2 MCQs)

 

7.6 Integration by Parts

This part would teach you methods on how to integrate products of functions. This method is also called partial integration. If for a single variable a, there are 2 differentiable functions f1 and f2, then by applying product rule of differentiation we can write d (f1f2)/da = f1df2/da + f2 df1/da.

You would then learn how to apply integration on both sides and conclude that integral of the product of 2 functions is given by:

  • (1st function) * (integral of the 2nd function) – Integral of ((differential coefficient of 1st function) * (integral of the 2nd function))

  • Exercise 7.6 Solutions: 24 Questions (22 Short Questions, 2 MCQ)

  • Exercise 7.7 Solutions: 11 Questions (9 Short Questions, 2 MCQ)

 

7.7 Definite Integral

In this section, you will closely learn about what a definite integral is. You would learn how to denote a definite integral and its notation;

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx

In this, a is the lower limit and b is the upper limit of the integral.

 

You would also learn how to represent a definite integral either as a limit of a sum or an antiderivative in the interval (a, b).

 

7.8 Fundamental Theorem of Calculus

In this chapter, you would learn how concepts of integration and differentiation of a function are linked. This is done by calculating the anti-derivative difference at the upper and lower limits of the integration process.

 

You Would Learn The Following

Area Function -

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx
    defines the area of the curve y = f(x) which is bounded by ordinates of the x line (a and b). So, we can say that the area A(x) =

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx

 

First Fundamental Theorem of Integral Calculus – A’(x) = f(x) for all x ε

  • a,b

  • a,b, where f is a continuous function on the closed interval

  • a,b

  • a,b.

 

2nd Fundamental Theorem of Integral Calculus -

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx
    =

  • F(x)

  • F(x)ba = F(b) – F(a), here f - is a continuous function on the closed interval

  • a,b

  • a,b

  • F – antiderivative of f.

  • Exercise 7.9 Solutions: 22 Questions (20 Short Questions, 2 MCQs)

 

7.9 Evaluation of Definite Integrals by Substitution

In this chapter, you would learn steps to evaluate definite integrals

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx

, by substitution as described below:

  • Reduce the given integral to a known form by replacing y = f(x) and x = q(y) (considering the integral without limits).

  • Now integrate the new integrand without mentioning the constant of integration.

  • Exercise 7.10 Solutions: 10 Questions (8 Short Questions, 2 MCQs)

 

7.10 Some Properties of Definite Integral

Here you will learn many properties of definite integrals which will simplify the process of evaluating definite integrals. Some of the properties mentioned are as below:

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx
    =

  • b

  • a

  • f(t)dt

  • ∫abf(t)dt

  • 2p

  • 0

  • f(x)dx

  • ∫02pf(x)dx
    =

  • p

  • 0

  • f(x)dx

  • ∫0pf(x)dx
    +

  • p

  • o

  • f(2p−x)dx

  • ∫opf(2p−x)dx

  • p

  • 0

  • f(x)dx

  • ∫0pf(x)dx
    =

  • p

  • 0

  • f(p−x)dx

  • ∫0pf(p−x)dx

  • b

  • a

  • f(x)dx

  • ∫abf(x)dx
    =

  • p

  • a

  • f(x)dx

  • ∫apf(x)dx
    +

  • b

  • p

  • f(x)dx

  • ∫pbf(x)dx

You will also learn how to prove all these properties in this section of Integrals Class 12 Notes Maths Chapter 7.

  • Exercise 7.11 Solutions: 21 Questions (19 Short Questions, 2 MCQs)

 

Key Features of NCERT Solutions for Class 12 Maths Chapter 7

Maths is a difficult subject for many students and sometimes you might struggle in finding the right solution. That is why a ready-made solution by experienced teachers is a valuable resource for anyone who wants to master the subject and also prepare for various prestigious competitive exams. You will benefit immensely from these Maths NCERT Solutions Class 12 Chapter 7 because:

 

  • Comprehensive explanations for each exercise and questions, promote a deeper understanding of the subject.

  • Clear and structured presentation for easy comprehension.

  • Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

  • Visual aids like diagrams and illustrations to simplify complex concepts.

  • Additional tips and insights to enhance students' performance.

  • Chapter summaries for quick revision.

  • Online accessibility and downloadable resources for flexible study and revision.


Why Choose Vedantu’s NCERT Solutions for Class 12 Maths? 

These NCERT Solutions curated by Vedantu are aimed at mastering the concepts and acquiring comprehensive knowledge about the various types of sums on Integrals that can be expected in examinations. All questions covered in NCERT Maths Class 12 Chapter 7 are provided with appropriate solutions on Vedantu, which will come in handy while revising the CBSE Class 12 Mathematics syllabus. 

 

NCERT Solutions for Maths Class 12  Chapter 7 available in PDF format are designed by our subject experts, as per the latest CBSE curriculum. These solutions are helpful for the students to streamline their last-minute revisions. Students can download the PDFs from Vedantu for free and study offline anytime and from anywhere.

 

We, at Vedantu, understand that your life as a student has many challenges. From assignments to classes to other activities outside school, you need to cope up with all that. Hence finding Integrals Class 12 NCERT solutions PDF online is going to give you a much-needed break from your hectic life.

 

The official website of Vedantu now has Class 12 Maths NCERT Solutions Chapter 7 which can be downloaded anytime anywhere and quickly revise all important points of problems of Chapter 7 Maths Class 12.

 

Give your preparation a boost by solving all the NCERT Solutions of Class 12 Maths Chapter 7 Integrals by referring to the free PDF available on our website. Prepared by subject experts, these detailed and step-by-step solutions will guide you how to answer a Maths problem.


We have also provided a summary of the chapter and important topics covered in each exercise so that you can get everything you need at one place only. So, enhance your performance and fetch high marks with the right preparation material by downloading the PDF today!

 

Conclusion

Vedantu's NCERT Solutions for Class 12 Maths Chapter 7 - Integrals offer a transformative learning experience. Navigating complex problems becomes accessible, thanks to user-friendly solutions, coupled with continuous support for doubt resolution. This platform not only simplifies integrals but also ensures a comprehensive understanding of mathematical concepts. With Vedantu's commitment to accessible and supportive learning, mastering Class 12 Maths becomes an achievable goal. Empower your mathematical journey, conquer integrals, and excel in your academic pursuits with the reliable guidance of Vedantu's NCERT Solutions.


NCERT Solutions for Class 12 Maths


You will also understand how to prove these standard integrals, some of those are mentioned below:

  • ∫ dx/ (x2 – b2) = 1/2b log |x – b/x + a| + C

  • ∫ dx/ (x2 + b2) = 1/b tan-1 x/b + C

  • ∫ dx/ (√x2 + b2) = log |x + √x2 + b2| + C

Exercise 7.4 Solutions: 25 Questions (23 Short Questions, 2 MCQs)


7.5 Integration by Partial Functions

This unit of Chapter 7 Class 12 Maths NCERT solutions further expands what you learned in unit 7.3 about integration by partial functions. In this section, you would learn how an improper rational function can be converted into a proper rational function by a long division process. An improper rational function is the one where for 2 polynomials p1(x) and p2(x) on x,  p2(x) <>0 and degrees of p1(x) > p2(x) t.


You will learn here how to disintegrate an equation into parts and find the integral of each of the parts. This is known as partial fraction decomposition. 


Exercise 7.5 Solutions: 23 Questions (21 Short Questions, 2 MCQs)


7.6 Integration by Parts

This part would teach you methods on how to integrate products of functions. This method is also called partial integration. If for a single variable a, there are 2 differentiable functions f1 and f2, then by applying product rule of differentiation we can write d (f1f2)/da = f1df2/da + f2 df1/da.


You would then learn how to apply integration on both sides and conclude that integral of the product of 2 functions is given by:


(1st function) * (integral of the 2nd function) – Integral of ((differential coefficient of 1st function) * (integral of the 2nd function))


Exercise 7.6 Solutions: 24 Questions (22 Short Questions, 2 MCQ)


Exercise 7.7 Solutions: 11 Questions (9 Short Questions, 2 MCQ)


7.7 Definite Integral

In this section, you will closely learn about what a definite integral is. You would learn how to denote a definite integral and its notation; \[\int_{a}^{b}f(x)dx\] . In this, a is the lower limit and b is the upper limit of the integral.


You would also learn how to represent a definite integral either as a limit of a sum or an antiderivative in the interval (a, b). 


Exercise 7.8 Solutions: 6 Questions (6 Short Questions)


7.8 Fundamental Theorem of Calculus

In this chapter, you would learn how concepts of integration and differentiation of a function are linked. This is done by calculating the anti-derivative difference at the upper and lower limits of the integration process. 

You Would Learn The Following

  • Area Function - \[\int_{a}^{b}f(x)dx\] defines the area of the curve y = f(x) which is bounded by ordinates of the x line (a and b). So, we can say that the area A(x) = \[\int_{a}^{b}f(x)dx\]

  • First Fundamental Theorem of Integral Calculus – A’(x) = f(x) for all x ε [a,b], where f is a continuous function on the closed interval [a,b].

  • 2nd Fundamental Theorem of Integral Calculus - \[\int_{a}^{b}f(x)dx\] = [F(x)]ba = F(b) – F(a), here f - is a continuous function on the closed interval [a,b]

          F – antiderivative of f.

Exercise 7.9 Solutions: 22 Questions (20 Short Questions, 2 MCQs)


7.9 Evaluation of Definite Integrals by Substitution

In this chapter, you would learn steps to evaluate definite integrals \[\int_{a}^{b}f(x)dx\] , by substitution as described below:

  • Reduce the given integral to a known form by replacing y = f(x) and x = q(y) (considering the integral without limits).

  • Now integrate the new integrand without mentioning the constant of integration.

Exercise 7.10 Solutions: 10 Questions (8 Short Questions, 2 MCQs)


7.10 Some Properties of Definite Integral

Here you will learn many properties of definite integral which will simplify the process of evaluating definite integrals. Some of the properties mentioned are as below:

  • \[\int_{a}^{b}f(x)dx\] = \[\int_{a}^{b}f(t)dt\]

  • \[\int_{0}^{2p}f(x)dx\] = \[\int_{0}^{p}f(x)dx\] + \[\int_{o}^{p}f(2p-x)dx\]

  • \[\int_{0}^{p}f(x)dx\] = \[\int_{0}^{p}f(p-x)dx\]

  • \[\int_{a}^{b}f(x)dx\] = \[\int_{a}^{p}f(x)dx\] + \[\int_{p}^{b}f(x)dx\]

You will also learn how to prove all these properties in this section of Integrals Class 12 Notes Maths Chapter 7.

Exercise 7.11 Solutions: 21 Questions (19 Short Questions, 2 MCQs)


Key Features of NCERT Solutions for Class 12 Maths Chapter 7

Maths is a difficult subject for many students and sometimes you might struggle in finding the right solution. That is why a readymade solution by experienced teachers is a valuable resource for anyone who wants to master the subject and also prepare for various prestigious competitive exams. You will benefit immensely from these Maths NCERT Solutions Class 12 Chapter 7 because:

  • You get to know the right way of cracking difficult sums by a team of experts who will provide the best way of approaching any problem.

  • The solutions would be a quick way of revising all the major formulas mentioned in the chapter.

  • The online solutions can be accessed anytime, and our teachers will guide if you need it.

  • Solutions will help boost your confidence and check your preparedness for exams


NCERT Solutions for Class 12 Maths

FAQs on NCERT Solutions for Class 12 Maths Chapter 7 - Integrals

1. What is integral in simple terms?

In Calculus, an integral is often referred to as the area under a curve. As mathematical functions can be represented on a graph paper, the area enclosed between the curve of a function and the x-axis is the integral value of that particular equation. You will come across a list of formulas for calculating the integration of various functions, in class 12 calculus. Integral is also known as anti-derivative and you may observe that the integration formulas for some functions are just the reverse of that of their differentiation formulas. Integral calculus is one of the most important topics of Class 12 Mathematics.

2. What is the difference between definite and indefinite integrals?

When the upper and lower limits are given for an integral, it is called a definite integral. While calculating the definite integration of a given function f(x), for upper limit a, and lower limit b, it must be noted that you are calculating the area under the curve f(x) from x=a to x=b. Unlike definite integrals, upper and lower limits are not provided for indefinite integrals. So, you calculate a generic integral value for a family of similar functions f(x), in indefinite integral, where ‘x’ can have a range of solutions. Also, there is a notation for additive constant, C, written along with the indefinite integral value of any function.

3. Is Class 12 integral calculus difficult to understand?

No, Class 12 integral calculus is not much difficult to understand, instead it is one of the most interesting topics in the Maths syllabus 12th boards. The basic sums of integration can be solved if you have a good knowledge of the integration formulas for various types of functions. Also, understanding the concepts of integral calculus becomes easier if you have proper knowledge of derivatives. Integration is also termed as the antiderivative. A good understanding of all the concepts of integral calculus introduced in Class 12 is very essential. These concepts lay the foundation of the theories of advanced mathematics and statistical studies.

4. Where can I find reliable NCERT Solutions for Class 12 Maths Chapter 7- Integrals online?

You can find reliable NCERT Solutions for Class 12 Maths Chapter 7 - Integrals on Vedantu. The NCERT solutions on Vedantu are prepared by our team of subject matter experts. All the sums of NCERT Class 12 Maths Chapter 7 are worked out in a stepwise manner so that students can verify their own solutions. Our subject matter experts have followed the updated CBSE guidelines for Class 12, for preparing the NCERT Maths solutions for integrals. You can download these NCERT solutions for free and you can refer to them for practice purposes.

5. What is the meaning of integrals in NCERT Class 12 Maths Chapter 7?

Integrals in Mathematical language can be described as a generalization of area or area under a curve of a function. They come under the topic of calculus along with derivatives. It can also be represented as a numerical value. Learning about integrals and derivatives can be fun and interesting but at the same time, thorough practice is needed as integrals are often very tricky and there can be multiple ways of solving the same problem. 

6. Where can I download the NCERT Solutions for Class 12 Maths Chapter 7 PDF online?

To download the solutions for Class 12 Maths Chapter 7, follow the steps mentioned below:

  • Visit the page-NCERT Solutions for Class 12 Maths.

  • Choose the chapter of your choice, here, it is Chapter 7

  • The webpage with Vedantu’s solutions for Class 12 Maths Chapter 7 will open.

  • To download this, click on the Download PDF button and you can view the solutions offline free of cost.

For more efficient doubt clearing sessions or other help, one can visit the Vedantu site and avail their help for cracking exams and achieving good marks. The students are advised to first try out the exercises based on their knowledge and practice before referring to the solution so that they can get a clear idea as to where they are lacking behind.

7. Why are integrals so hard?

If any student skips any topic or finds it hard to grasp the basic idea of any topic, integrals could be a nightmare for that student. Hence, it is crucial that you practice all different types of sums and get used to the different difficulty levels of the questions. Practice as many sums as you can to completely understand any topic.

8. What do you mean by differentiation?

Differentiation is a technique for determining a function's derivative. Differentiation is a mathematical procedure for determining the instantaneous rate of change of a function depending on one of its variables. The most prominent example is velocity, which is the rate of change of displacement with respect to time. It is an important topic and hence must be practised judiciously.

9. What are definite and indefinite integrals?

In cases when the lower and upper bounds are constants, the number can be represented or solved by a definite integral. There are several functions whose derivatives are represented by the indefinite integral, and they are called f's. Any two functions in the same family will always be distinct from each other. For a detailed explanation of the same, you can visit Vedantu website and the app.