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Important Questions for CBSE Class 12 Maths Chapter 7- Integrals 2023-24

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CBSE Class 12 Maths Chapter-7 Important Questions - Free PDF Download

In order to help the students in their exam preparations, Class 12 Maths Integration Important Questions were developed by subject experts. In Class 12 Maths Chapter 7 Extra Questions contains the idea of integrals. Students learn about integral calculus (definite and indefinite), its properties, and much more in this chapter. For both the CBSE board exam and competitive examinations, this subject is extremely relevant. In this chapter, the notions of integrals are given in a thorough and easy to understand way. These Important Questions are very simple and can very easily assist students in learning the process of problem-solving. 

In students, these questions build a solid conceptual base that plays a major role in successfully preparing for the competitive exams in the later stages. We provide comprehensive answers to issues where students can easily prepare in a much better and productive way all the topics covered in the syllabus of their respective classes and to crack the toughest competitive examinations. Students searching for questions in NCERT's books can now put a stop to their search and get them here in one place.  

According to the new NCERT guidelines, Integration Important Questions Class 12 provides you with the latest questions and solutions. In order to support you better, students can explain all their doubts about each chapter by practising these important chapter questions and thorough explanations given by our experts. Due to time constraints, these questions will help students prepare well for the exams. The chapter-wise strategy for preparing for your board examination offers you Integration Important Questions Class 12.


Download CBSE Class 12 Maths Important Questions 2023-24 PDF

Also, check CBSE Class 12 Maths Important Questions for other chapters:

CBSE Class 12 Maths Important Questions

Sl.No

Chapter No

Chapter Name

1

Chapter 1

Relations and Functions

2

Chapter 2

Inverse Trigonometric Functions

3

Chapter 3

Matrices

4

Chapter 4

Determinants

5

Chapter 5

Continuity and Differentiability

6

Chapter 6

Application of Derivatives

7

Chapter 7

Integrals

8

Chapter 8

Application of Integrals

9

Chapter 9

Differential Equations

10

Chapter 10

Vector Algebra

11

Chapter 11

Three Dimensional Geometry

12

Chapter 12

Linear Programming

13

Chapter 13

Probability

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Study Important Questions Class 12 Maths Chapter -7 Integrals

Very Short Answer Type Questions: (1 Mark)

Evaluate the following integrals 

1. \[\int{\left( {{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x} \right)}dx\]

Ans: We know that ${{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}$ is $\frac{\pi }{2}$.

This implies $\int{\left( {{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x} \right)}dx=\int{\frac{\pi }{2}dx}$

Solving we get,

$\int{\frac{\pi }{2}dx}=\frac{\pi }{2}x+C$.


2. $\int\limits_{-1}^{1}{{{e}^{\left| x \right|}}dx}$

Ans: Let $I=\int\limits_{-1}^{+1}{{{e}^{\left| x \right|}}dx}$

Now, $f\left( x \right)={{e}^{\left| x \right|}}$

$f\left( -x \right)={{e}^{\left| -x \right|}}={{e}^{\left| x \right|}}=f\left( x \right)$

So, $f\left( x \right)$ is an even function.

$\begin{align} & \int\limits_{-1}^{+1}{{{e}^{\left| x \right|}}}dx=2\int\limits_{0}^{1}{{{e}^{\left| x \right|}}}dx \\ & =2\int\limits_{0}^{1}{{{e}^{x}}dx}\left( as,\left| x \right|=x,x>0 \right) \\ \end{align}$

\[\begin{align} & =2\left[ {{e}^{x}} \right]_{0}^{1} \\ & =2\left( e-1 \right) \\ & =2e-2 \\ \end{align}\]


3. $\int{\frac{1}{1-{{\sin }^{2}}x}dx}$

Ans: 

$\begin{align} & \int{\frac{1}{1-{{\sin }^{2}}x}dx} \\ & =\int{\frac{1}{{{\cos }^{2}}x}dx} \\ & =\int{{{\sec }^{2}}xdx} \\ & =\tan x+C \\ \end{align}$


4. $\int{\left( {{8}^{x}}+{{x}^{8}}+\frac{8}{x}+\frac{x}{8} \right)}dx$

Ans: 

$\begin{align} & \int{{{8}^{x}}dx+\int{{{x}^{8}}dx}}+\int{\frac{8}{x}dx+\int{\frac{x}{8}}dx} \\ & =\frac{{{8}^{x}}}{\log 8}+\frac{{{x}^{9}}}{9}+8\log \left| x \right|+\frac{{{x}^{2}}}{16} \\ \end{align}$


5. $\int\limits_{-1}^{1}{{{x}^{99}}{{\cos }^{4}}xdx}$

Ans: $\begin{align} & f\left( x \right)={{x}^{99}}{{\cos }^{4}}x \\ & f\left( -x \right)={{\left( -x \right)}^{99}}{{\left( \cos \left( -x \right) \right)}^{4}} \\ & =-{{x}^{99}}{{\cos }^{4}}x=-f\left( x \right) \\ \end{align}$

We know that property \[\int\limits_{-a}^{a}{f\left( x \right)}dx=0\], when $f\left( -x \right)=-f\left( x \right)$

So, $\int\limits_{-1}^{1}{{{x}^{99}}{{\cos }^{4}}xdx=0}$


6. $\int{\frac{1}{x\log x\log \left( \log x \right)}dx}$

Ans: 

\[\begin{align} & I=\int{\frac{1}{x\log x\log \left( \log x \right)}dx} \\ & I=\int{\frac{1}{\log \left( \log x \right)}\left\{ \frac{dx}{x.\log \left( x \right)} \right\}} \\ \end{align}\]

Let $\log \left( \log x \right)=t$

$\begin{align} & \Rightarrow \frac{1}{\log x}.\frac{1}{x}.dx=dt \\ & \Rightarrow \frac{dx}{x.\log \left( x \right)}=dt \\ \end{align}$

Hence,

 $\begin{align} & I=\int{\frac{dt}{t}} \\ & =\log \left| t \right|+C \\ & =\log \left| \log \left( \log x \right) \right|+C \\ \end{align}$


7. $\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{4+3\sin x}{4+3\cos x} \right)dx}$

Ans:$I=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{4+3\sin x}{4+3\cos x} \right)}.....\left( 1 \right)$

 $\begin{align} & I=\int\limits_{0}^{\frac{\pi }{2}}{\log \left[ \frac{4+3\sin \left( \frac{\pi }{2}-x \right)}{4+3\cos \left( \frac{\pi }{2}-x \right)} \right]}dx \\ & \Rightarrow I=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{4+3\cos x}{4+3\sin x} \right)}dx.....\left( 2 \right) \\ \end{align}$

Adding (1) and (2),

$\begin{align} & 2I=\int\limits_{0}^{\frac{\pi }{2}}{\log \left[ \left( \frac{4+3\sin x}{4+3\cos x} \right)\times \left( \frac{4+3\cos x}{4+3\sin x} \right) \right]}dx \\ & =\int\limits_{0}^{\frac{\pi }{2}}{\left( \log 1 \right)dx} \\ & =\int\limits_{0}^{\frac{\pi }{2}}{\left( 0 \right)dx} \\ & =0 \\ \end{align}$


8. $\int{\left( {{e}^{a\log x}}+{{e}^{x\log a}} \right)dx}$

Ans: We know that 

$\begin{align} & a\log x=\ln {{x}^{a}} \\ & x\log a=\ln {{a}^{x}} \\ \end{align}$

$\int{\left( {{e}^{a\log x}}+{{e}^{x\log a}} \right)dx}=\int{\left( {{e}^{\ln {{x}^{a}}}}+{{e}^{\ln {{a}^{x}}}} \right)}dx$

Also, we know that ${{e}^{\ln \left( f\left( x \right) \right)}}=f\left( x \right)$

$\begin{align} & =f\left( {{x}^{a}}+{{a}^{x}} \right)dx \\ & =\frac{{{x}^{a+1}}}{a+1}+\frac{{{a}^{x}}}{\ln a}+c \\ \end{align}$


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

Ans: We know that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$

$\begin{align} & \int{\left( \frac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)dx=\int{\left( \frac{{{\cos }^{2}}x-{{\sin }^{2}}x+2{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)}} \\ & =\int{\left( \frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)dx=\int{\left( \frac{1}{{{\cos }^{2}}x} \right)dx}} \\ \end{align}$

(As ${{\cos }^{2}}x+{{\sin }^{2}}x=1$)

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


10. $\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{{{\sin }^{7}}x}dx$

Ans: Let $f\left( x \right)={{\sin }^{7}}x$

$f\left( -x \right)={{\sin }^{7}}\left( -x \right)=-{{\sin }^{7}}x$

This implies that f(x) is an odd function.

$\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{{{\sin }^{7}}xdx=0}$

($\because \int\limits_{-a}^{a}{f\left( x \right)dx=0}$ if f(x) is an odd function).


11. $\int{\left( {{x}^{c}}+{{c}^{x}} \right)dx}$

Ans: Using the properties,

$\int{{{x}^{n}}dx}=\frac{{{x}^{n+1}}}{n+1}+C,\int{{{n}^{x}}dx=\frac{{{n}^{x}}}{\ln n}+C}$ 

$\int{\left( {{x}^{c}}+{{c}^{x}} \right)dx=\frac{{{x}^{c+1}}}{c+1}}+\frac{{{c}^{x}}}{\ln c}+C$


12. $\frac{d}{dx}\left[ \int{f\left( x \right)dx} \right]$

Ans: Differentiating and integrating a function at a time leaves the function itself. So,

$\frac{d}{dx}\left[ \int{f\left( x \right)dx} \right]=f\left( x \right)+C$


13. $\int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

Ans: $\int{\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx=\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}}$ (Using ${{\cos }^{2}}x+{{\sin }^{2}}x=1$)

$\begin{align} & =\int{\frac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx+\int{\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx} \\ & =\int{\frac{1}{{{\cos }^{2}}x}dx+\int{\frac{1}{{{\sin }^{2}}x}dx}} \\ & =\int{{{\sec }^{2}}xdx+\int{\cos e{{c}^{2}}xdx}} \\ & =\tan x-\cot x+C \\ \end{align}$


14. $\int{\frac{1}{\sqrt{x}+\sqrt{x-1}}dx}$

Ans: Rationalizing we have,

$\begin{align} & \int{\frac{1}{\sqrt{x}+\sqrt{x-1}}\times \frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x}-\sqrt{x-1}}dx} \\ & =\int{\frac{\sqrt{x}-\sqrt{x-1}}{x-x+1}}dx \\ & =\int{\sqrt{x}-\sqrt{x-1}}dx \\ & =\frac{{{x}^{\frac{3}{2}}}}{3/2}-\frac{{{\left( x-1 \right)}^{\frac{3}{2}}}}{3/2}+C \\ & =\frac{2}{3}{{x}^{\frac{3}{2}}}-\frac{2}{3}{{\left( x-1 \right)}^{\frac{3}{2}}}+C \\ \end{align}$


15. $\int{{{e}^{-\log {{e}^{x}}}}dx}$

Ans: We know that $\log {{e}^{x}}=x$

Hence, $\int{{{e}^{-\log {{e}^{x}}}}dx}=\int{{{e}^{-x}}dx}$

Now, put $u=-x$

 $\begin{align} & \frac{du}{dx}=-1 \\ & \Rightarrow dx=-du \\ \end{align}$

$\begin{align} & \int{{{e}^{-x}}}dx=\int{-{{e}^{u}}}du \\ & =-\int{{{e}^{u}}}du=-{{e}^{u}}+C \\ & =-{{e}^{-x}}+C \\ \end{align}$


16. $\int{\frac{{{e}^{x}}}{{{a}^{x}}}dx}$

Ans: 

$\begin{align} & \int{\frac{{{e}^{x}}}{{{a}^{x}}}dx=\int{{{\left( \frac{e}{a} \right)}^{x}}dx}} \\ & =\frac{{{\left( \frac{e}{a} \right)}^{x}}}{\log \left( \frac{e}{a} \right)}+C \\ & \left( \int{{{m}^{x}}=\frac{{{m}^{x}}}{\log m}+C} \right) \\ \end{align}$


17. $\int{{{2}^{x}}{{e}^{x}}dx}$

Ans: 

$\begin{align} & \int{{{2}^{x}}{{e}^{x}}}dx=\int{{{\left( 2e \right)}^{x}}}dx \\ & =\frac{1}{\log \left( 2e \right)}.{{\left( 2e \right)}^{x}}+C \\ & =\frac{{{\left( 2e \right)}^{x}}}{\log \left( 2e \right)}+C \\ & =\frac{{{2}^{x}}{{e}^{x}}}{\log \left( 2e \right)}+C \\ \end{align}$


18. $\int{\frac{x}{\sqrt{x+1}}dx}$

Ans: Let $u=x+1,du=dx$

$\Rightarrow u-1=x$

$\begin{align} & \int{\frac{u-1}{\sqrt{u}}du}=\int{\frac{u}{\sqrt{u}}du-\int{\frac{1}{\sqrt{u}}du}} \\ & =\int{\frac{u}{{{u}^{\frac{1}{2}}}}du-}\int{\frac{1}{{{u}^{\frac{1}{2}}}}du} \\ & =\int{{{u}^{\frac{1}{2}}}}du-\int{{{u}^{-\frac{1}{2}}}}du \\ & =\frac{{{u}^{\frac{1}{2}+1}}}{\frac{1}{2}+1}-\frac{{{u}^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1}+C \\ & =\frac{{{u}^{\frac{3}{2}}}}{3/2}-\frac{{{u}^{\frac{1}{2}}}}{1/2}+C \\ & =\frac{2}{3}{{u}^{\frac{3}{2}}}-2{{u}^{\frac{1}{2}}}+C \\ \end{align}$

Substituting the value of u we have,

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


19. $\int{\frac{x}{{{\left( x+1 \right)}^{2}}}dx}$

Ans: Let us assume that $u=x+1,du=dx$

$\Rightarrow u-1=x$

Now, substituting in the given integral we get,

$\int{\frac{u-1}{{{u}^{2}}}du=\int{\left( \frac{1}{u}+\frac{1}{{{u}^{2}}} \right)du}}$

Now, by using the power formula of integration we get,

$\begin{align} & =\ln \left( u \right)+\frac{{{u}^{-2+1}}}{-2+1}+c \\ & =\ln \left( u \right)+\frac{{{u}^{-1}}}{-1}+c \\ & =\ln \left( u \right)-\frac{1}{u}+c \\ \end{align}$

Now replacing the value of $\text{u}$ we get,

$=\ln \left( x+1 \right)-\frac{1}{x+1}+c$


20. $\int{\frac{{{e}^{\sqrt{x}}}}{x}}dx$

Ans: Let us assume $u=\sqrt{x}\Rightarrow {{u}^{2}}=x$

Now differentiating on both sides,

$\text{2udu}=\text{dx}$

Now, substituting in the given expression we get,

$\begin{align} & =\int{\frac{{{e}^{u}}}{{{u}^{2}}}\left( 2udu \right)} \\ & =2.\int{\frac{{{e}^{u}}}{u}}du \\ \end{align}$

By using the formula $\int{\frac{{{e}^{x}}}{x}dx}=Ei\left( x \right)$ we get,

$=2Ei\left( u \right)+c$

Substituting back in terms of $\text{x}$ we have,

$=2Ei\left( \sqrt{x} \right)+c$


21. $\int{{{\cos }^{2}}\alpha }dx$

Ans: We know that ${{\cos }^{2}}\alpha $ is a constant.

So, $\int{{{\cos }^{2}}\alpha }dx=x{{\cos }^{2}}\alpha +C$


22. \[\int{\frac{1}{x\cos \alpha +1}dx}\]

Ans: $I=\int{\frac{1}{x\cos \alpha +1}dx}$

Let $x\cos \alpha +1=t$ 

$\begin{align} & \cos \alpha dx=dt \\ & dx=\frac{dt}{\cos \alpha } \\ \end{align}$

$\begin{align} & I=\int{\frac{1}{t}.\frac{dt}{\cos \alpha }} \\ & I=\frac{1}{\cos \alpha }\int{\frac{dt}{t}}=\frac{1}{\cos \alpha }\log \left| t \right|+C \\ & I=\frac{\log \left| x\cos \alpha +1 \right|}{\cos \alpha }+C \\ \end{align}$


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

Ans: Substitute $u=\log \left( \sec x+\tan x \right)$

$du=\frac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x}dx\And dx=\frac{\sec x+\tan x}{\sec x\tan x+{{\sec }^{2}}x}du$

$\begin{align} & I=\int{\sec x.u.\frac{\sec x+\tan x}{\sec x\tan x+{{\sec }^{2}}x}}du \\ & I=\int{\sec x.u.\frac{\sec x+\tan x}{\sec x\left( \tan x+\sec x \right)}}du \\ & I=\int{u}du \\ & I=\frac{1}{2}{{u}^{2}}+C \\ & I=\frac{1}{2}{{\log }^{2}}\left( \sec x+\tan x \right)+C \\ & I=\frac{{{\left( \log \left| \sec x+\tan x \right| \right)}^{2}}}{2}+C \\ \end{align}$


24. $\int{\frac{1}{\cos \alpha +x\sin \alpha }}dx$

Ans: Substitute \[u=\cos \alpha +x\sin \alpha \]

$\begin{align} & \int{\frac{1}{\cos \alpha +x\sin \alpha }}dx=\int{\frac{1}{u\sin \alpha }}du \\ & =\frac{1}{\sin \alpha }.\int{\frac{1}{u}}du\left[ \int{a.f\left( x \right)dx=a.\int{f\left( x \right)dx}} \right] \\ \end{align}$

Now, use $\int{\frac{1}{u}du}=\ln \left( \left| u \right| \right)$

\[\begin{align} & =\frac{1}{\sin \alpha }\log \left| u \right| \\ & =\frac{1}{\sin \alpha }\log \left| \cos \alpha +x\sin \alpha \right|+C \\ \end{align}\]


25. $\int{\cot x.\log \sin x}dx$

Ans: We know the formula of cotangent ratio as,

$\text{cotx}=\frac{\text{cosx}}{\text{sinx}}$

By using the above formula for given expression we get,

\[\int{\frac{\cos x\log \left( \sin x \right)}{\sin x}dx}\]

Substitute $u=\sin x$

Now, differentiating on both sides we get,

$\text{du}=\text{cosxdx}$

Now substituting in the given expression we get,

$=\int{\frac{\log u}{u}}du$

Substitute $v=\log u$ so that by differentiating we have,

$\text{dv}=\frac{\text{1}}{\text{u}}\text{du}$

Now substituting in the above integral we get,

$\begin{align} & =\int{v}dv \\ & =\frac{{{v}^{2}}}{2} \\ & =\frac{{{\log }^{2}}\left( \sin x \right)}{2} \\ & =\frac{{{\left( \log \sin x \right)}^{2}}}{2}+C \\ \end{align}$


26. $\int{{{\left( x-\frac{1}{2} \right)}^{3}}}dx$

Ans:

 \[\begin{align} & \int{{{\left( x-\frac{1}{2} \right)}^{3}}dx}=\int{\left( {{x}^{3}}-\frac{3{{x}^{2}}}{2}+\frac{3x}{4}-\frac{1}{8} \right)}dx \\ & =\int{{{x}^{3}}}dx-\int{\frac{3{{x}^{2}}}{2}}dx+\int{\frac{3x}{4}}dx-\int{\frac{1}{8}}dx \\ & =\frac{{{x}^{4}}}{4}-\frac{{{x}^{3}}}{2}+\frac{3{{x}^{2}}}{8}-\frac{1}{8}x+c \\ \end{align}\]


27. $\int{\frac{1}{x\left( 2+3\log x \right)}}dx$

Ans: Substitute $u=2+3\log x$

$\begin{align} & \int{\frac{1}{x\left( 2+3\log x \right)}dx=\int{\frac{1}{3u}}du} \\ & =\frac{1}{3}\int{\frac{1}{u}}du \\ & =\frac{1}{3}\log u\left[ \int{\frac{1}{u}}du=\log u \right] \\ & =\frac{1}{3}\log \left| 2+3\log x \right|+C \\ \end{align}$


28. $\int{\frac{1-\sin x}{x+\cos x}}dx$

Ans: Substitute $u=x+\cos x$

$\begin{align} & \int{\frac{1-\sin x}{x+\cos x}dx}=\int{\frac{1}{u}}du \\ & =\log u \\ & =\log \left| x+\cos x \right|+C \\ \end{align}$


29. $\int{\frac{1-\cos x}{\sin x}}dx$

Ans: Separating each term in the numerator with the denominator of the expression $\frac{1-\cos x}{\sin x}$ we get,

$\int{\frac{1-\cos x}{\sin x}dx=\int{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}dx}$

By sum rule,

$=\int{\frac{1}{\sin x}dx-\int{\frac{\cos x}{\sin x}}}dx$

$=\int{\csc xdx-\int{\cot x}}dx$

By using the standard results of differentiation we get,

$=\log \left| \tan \frac{x}{2} \right|-\log \left| \sin x \right|+C$


30. $\int{\frac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}dx}$

Ans: Let us take the given integral as,

 $\begin{align} & \int{\frac{{{x}^{e-1}}+{{e}^{x-1}}}{{{x}^{e}}+{{e}^{x}}}}dx=\int{\frac{{{x}^{e-1}}+\frac{{{e}^{x}}}{e}}{{{x}^{e}}+{{e}^{x}}}}dx \\ & =\int{\frac{e{{x}^{e-1}}+{{e}^{x}}}{e\left( {{x}^{e}}+{{e}^{x}} \right)}}dx \\ \end{align}$

Now, let us take, ${{\text{x}}^{\text{e}}}\text{+}{{\text{e}}^{\text{x}}}\text{=k}$, then differentiating on both sides we have,

$\left( \text{e}{{\text{x}}^{\text{e-1}}}\text{+}{{\text{e}}^{\text{x}}} \right)\text{dx}=\text{dk}$

So, we have the given expression as,

$\begin{align} & =\frac{1}{e}\int{\frac{dk}{k}} \\ & =\frac{1}{e}\ln \left( k \right)+c \\ & =\frac{1}{e}\ln \left( {{x}^{e}}+{{e}^{x}} \right)+c \\ \end{align}$


31. $\int{\frac{\left( x+1 \right)}{x}}\left( x+\log x \right)dx$

Ans: Substitute $u=x+\log x$

Differentiating both sides,

$\begin{align} & \frac{du}{dx}=1+\frac{1}{x} \\ & \left( 1+\frac{1}{x} \right)dx=du \\ \end{align}$

$\begin{align} & \int{\frac{\left( x+1 \right)}{x}\left( x+\log x \right)}dx=\int{u}dx \\ & \int{u}du=\frac{{{u}^{2}}}{2} \\ & =\frac{{{\left( x+\log x \right)}^{2}}}{2}+C \\ \end{align}$


32. $\int{{{\left( \sqrt{ax}-\frac{1}{\sqrt{ax}} \right)}^{2}}}dx$

Ans: Simplifying we have,

$\begin{align} & \int{{{\left( \sqrt{ax}-\frac{1}{\sqrt{ax}} \right)}^{2}}}dx=\int{ax-2+\frac{1}{ax}}dx \\ & =\int{ax}dx-\int{2}dx+\int{\frac{1}{ax}}dx \\ & =\frac{a{{x}^{2}}}{2}-2x+\frac{1}{a}\log \left| ax \right|+C \\ \end{align}$


33. $\int\limits_{0}^{\pi }{\left| \cos x \right|}dx$

Ans: $\int\limits_{0}^{\pi }{\left| \cos x \right|}dx=\int\limits_{0}^{\frac{\pi }{2}}{\cos x}dx+\int\limits_{\frac{\pi }{2}}^{\pi }{-\cos x}dx$

We know that 

$\begin{align} & \int\limits_{0}^{\frac{\pi }{2}}{\cos x}dx=1, \\ & \int\limits_{\frac{\pi }{2}}^{\pi }{-\cos x}dx=1 \\ & =1+1 \\ & =2 \\ \end{align}$


34. $\int\limits_{0}^{2}{\left[ x \right]}dx$ where [] is greatest integer function.

Ans: We know that when $\text{0}<\text{x}<\text{1}$ we have,

 $\Rightarrow \left[ x \right]=0$

Similarly when $\text{1}<\text{x}<\text{2}$ we have,

$\Rightarrow \left[ x \right]=1$

Therefore,

 $\begin{align} & \therefore \int\limits_{0}^{2}{\left[ x \right]}dx=\int\limits_{0}^{1}{\left[ x \right]}dx+\int\limits_{1}^{2}{\left[ x \right]}dx \\ & =\int\limits_{0}^{1}{0}dx+\int\limits_{1}^{2}{1}dx \\ & =0+\left( 2-1 \right) \\ & =1 \\ \end{align}$


35. $\int\limits_{0}^{\sqrt{2}}{\left[ {{x}^{2}} \right]}dx$ where [] is greatest integer function.

Ans: When

 $\begin{align} & \text{0}<\text{x}<\text{1} \\ & \Rightarrow \text{0}<{{\text{x}}^{\text{2}}}<\text{1} \\ & \Rightarrow \left[ {{\text{x}}^{\text{2}}} \right]\text{=0} \\ \end{align}$

Similarly, when

$\begin{align} & \text{1}<\text{x}<\sqrt{\text{2}} \\ & \text{1}<{{\text{x}}^{\text{2}}}<\text{2} \\ & \left[ {{\text{x}}^{\text{2}}} \right]=\text{1} \\ \end{align}$

Therefore, the integral is,

$\begin{align} & \therefore \int\limits_{0}^{\sqrt{2}}{\left[ {{x}^{2}} \right]}dx=\int\limits_{0}^{1}{\left[ {{x}^{2}} \right]}dx+\int\limits_{1}^{\sqrt{2}}{\left[ {{x}^{2}} \right]}dx \\ & =\int\limits_{0}^{1}{0}dx+\int\limits_{1}^{\sqrt{2}}{1}dx \\ & =0+\left( \sqrt{2}-1 \right) \\ & =\sqrt{2}-1 \\ \end{align}$


36. $\int\limits_{a}^{b}{\frac{f\left( x \right)}{f\left( x \right)+f\left( a+b-x \right)}}dx$

Ans: Let us assume the given integral as,

$\text{I}=\int\limits_{a}^{b}{\frac{\text{f}\left( \text{x} \right)}{\text{f}\left( \text{x} \right)\text{+f}\left( \text{a+b-x} \right)}\text{dx}}$……….(1)

We know the standard formula of differentiation as,

$\int\limits_{a}^{b}{\text{f}\left( \text{x} \right)\text{dx}}=\int\limits_{a}^{b}{\text{f}\left( \text{a+b-x} \right)\text{dx}}$

Using this formula for given integral we get,

$\text{I}=\int\limits_{a}^{b}{\frac{\text{f}\left( \text{a+b-x} \right)}{\text{f}\left( \text{a+b-x} \right)\text{+f}\left( \text{x} \right)}\text{dx}}$…………(2)

Now, adding both equations (1) and (2) w get,

$\begin{align} & \text{2I}=\int\limits_{a}^{b}{\frac{\text{f}\left( \text{x} \right)\text{+f}\left( \text{a+b-x} \right)}{\text{f}\left( \text{a+b-x} \right)\text{+f}\left( \text{x} \right)}\text{dx}} \\ & 2\text{I}=\int\limits_{a}^{b}{\text{dx}} \\ & \text{I}=\frac{\text{b}-\text{a}}{\text{2}} \\ \end{align}$


37. $\int\limits_{-2}^{1}{\frac{\left| x \right|}{x}}dx$

Ans: Eliminating absolutes we have,

$\int\limits_{-2}^{1}{\frac{\left| x \right|}{x}}dx=\int\limits_{-2}^{0}{\frac{-x}{x}}dx+\int\limits_{0}^{1}{\frac{x}{x}}dx$

On solving the above integrals we get,

$\begin{align} & \int\limits_{-2}^{0}{\frac{-x}{x}}dx=-2+\int\limits_{0}^{1}{\frac{x}{x}}dx=1 \\ & =-2+1 \\ & =-1 \\ \end{align}$


38. $\int\limits_{-1}^{1}{x\left| x \right|}dx$

Ans: Eliminating absolutes we have,

$\begin{align} & \int\limits_{-1}^{1}{x\left| x \right|}dx=\int\limits_{-1}^{0}{x\left( -x \right)dx+\int\limits_{0}^{1}{x.xdx}} \\ & Also, \\ & \int\limits_{-1}^{0}{x\left( -x \right)dx=-\frac{1}{3}}, \\ & \int\limits_{0}^{1}{x.xdx}=\frac{1}{3} \\ & =-\frac{1}{3}+\frac{1}{3} \\ & =0 \\ \end{align}$


39. If $\int\limits_{0}^{a}{\frac{1}{1+{{x}^{2}}}=\frac{\pi }{4}}$ then what is value of a.

Ans: 

$\begin{align} & \int{\frac{1}{1+{{x}^{2}}}=}{{\tan }^{-1}}x+C \\ & \int\limits_{0}^{a}{\frac{1}{1+{{x}^{2}}}={{\tan }^{-1}}}\left( a \right) \\ & {{\tan }^{-1}}\left( a \right)=\frac{\pi }{4} \\ & \tan \frac{\pi }{4}=a \\ & \Rightarrow a=1 \\ \end{align}$


40. $\int\limits_{a}^{b}{f\left( x \right)dx+\int\limits_{b}^{a}{f\left( x \right)}dx}$

Ans: Using the property $\int\limits_{p}^{q}{f\left( x \right)}dx=-\int\limits_{q}^{p}{f\left( x \right)}dx$ we have,

$\begin{align} & \int\limits_{a}^{b}{f\left( x \right)}dx+\int\limits_{b}^{a}{f\left( x \right)}dx=\int\limits_{a}^{b}{f\left( x \right)}dx-\int\limits_{a}^{b}{f\left( x \right)}dx \\ & =0 \\ \end{align}$


4 Mark Questions

41. Evaluate: 

(i). \[\int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}\]

Ans: - Let ${{\tan }^{-1}}{{x}^{2}}=t\Leftrightarrow \frac{dt}{dx}=\frac{2x}{1+{{x}^{4}}}$. Substitute the values in the given expression.

\[\int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}=\int{\csc \left( t \right)}\times \frac{dt}{2}\]

\[\Rightarrow \int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}=\frac{1}{2}\left[ \log \left| \csc \left( t \right)-\cot \left( t \right) \right| \right]+C\]

Substitute the value of $t$ in the above equation.

\[\begin{align} & \int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}=\frac{1}{2}\left[ \log \left| \csc \left( {{\tan }^{-1}}{{x}^{2}} \right)-\cot \left( {{\tan }^{-1}}{{x}^{2}} \right) \right| \right]+C \\ & \Rightarrow \int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}=\frac{1}{2}\left[ \log \left| \csc \left( {{\tan }^{-1}}{{x}^{2}} \right)-\frac{1}{{{x}^{2}}} \right| \right]+C \\ \end{align}\]

Hence the value of \[\int{\frac{x\csc \left( {{\tan }^{-1}}{{x}^{2}} \right)}{1+{{x}^{4}}}dx}\] is \[\frac{1}{2}\left[ \log \left| \csc \left( {{\tan }^{-1}}{{x}^{2}} \right)-\frac{1}{{{x}^{2}}} \right| \right]+C\].


(ii). $\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx}$.

Ans: - Rationalize the fraction by multiplying and dividing with $\sqrt{x+1}-\sqrt{x-1}$.

$\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx}=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\times \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}dx}$

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

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

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

$\Rightarrow \int{\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx}=\int{xdx}-\int{\sqrt{{{x}^{2}}-1}}dx$

Apply the integration formula $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx}=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C$ in the above equation.

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

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

Hence the value of $\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx}$ is $\frac{1}{2}\left[ {{x}^{2}}-x\sqrt{{{x}^{2}}-1}+\log \left| x+\sqrt{{{x}^{2}}-1} \right| \right]+C$.


(iii). $\int{\frac{1}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}$

Ans: - Write the given expression as 

$\int{\frac{1}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}=\frac{1}{\sin \left( a-b \right)}\int{\frac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}$

$\Rightarrow \int{\frac{1}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}=\frac{1}{\sin \left( a-b \right)}\int{\frac{\sin \left( x-b \right)\cos \left( x-a \right)-\sin \left( x-a \right)\cos \left( x-b \right)}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}$

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

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

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

Hence the value of $\int{\frac{1}{\sin \left( x-a \right)\sin \left( x-b \right)}dx}$ is $\frac{1}{\sin \left( a-b \right)}\log \left| \frac{\sin \left( x-a \right)}{\sin \left( x-b \right)} \right|+C$.


(iv). $\int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}$

Ans: - Modify the above expression as 

$\int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}=\int{\frac{\cos \left( x-a+2a \right)}{\cos \left( x-a \right)}}dx$

$\Rightarrow \int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}=\int{\frac{\cos \left( x-a \right)\cos 2a-\sin \left( x-a \right)\sin 2a}{\cos \left( x-a \right)}}dx$

$\Rightarrow \int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}=\int{\cos 2adx}-\sin 2a\int{\tan \left( x-a \right)dx}$

$\Rightarrow \int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}=x\cos 2a-\sin 2a\log \left| \sec \left( x-a \right) \right|+C$

Hence the value of $\int{\frac{\cos \left( x+a \right)}{\cos \left( x-a \right)}dx}$ is $x\cos 2a-\sin 2a\log \left| \sec \left( x-a \right) \right|+C$.


(v). $\int{\cos x\cos 2x\cos 3xdx}$

Ans: - Modify the given expression as 

$\int{\cos x\cos 2x\cos 3xdx}=\int{\frac{1}{2}\left( 2\cos x.\cos 3x \right)\cos 2xdx}$

Use the trigonometric formula $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ in the above equation.

$\int{\cos x\cos 2x\cos 3xdx}=\frac{1}{2}\int{\left( \cos 4x+\cos 2x \right)\cos 2xdx}$

$\Rightarrow \int{\cos x\cos 2x\cos 3xdx}=\frac{1}{2}\int{\left( \cos 4x.\cos 2x+{{\cos }^{2}}2x \right)}dx$

$\Rightarrow \int{\cos x\cos 2x\cos 3xdx}=\frac{1}{2}\int{\frac{1}{2}\left( 2\cos 4x.\cos 2x+2{{\cos }^{2}}2x \right)}dx$

Use the trigonometric formula $\cos 2\theta =2{{\cos }^{2}}\theta -1$ in the above equation.

\[\int{\cos x\cos 2x\cos 3xdx}=\frac{1}{4}\int{\left( \cos 6x+\cos 2x+\cos 4x+1 \right)dx}\]

$\Rightarrow \int{\cos x\cos 2x\cos 3xdx}=\frac{1}{4}\left[ x+\frac{\sin 2x}{2}+\frac{\sin 4x}{4}+\frac{\sin 6x}{6} \right]+C$

$\Rightarrow \int{\cos x\cos 2x\cos 3xdx}=\frac{1}{48}\left[ 12x+6\sin 2x+3\sin 4x+2\sin 6x \right]+C$

Hence the value of $\int{\cos x\cos 2x\cos 3xdx}$ is $\frac{1}{48}\left[ 12x+6\sin 2x+3\sin 4x+2\sin 6x \right]+C$.


(vi). $\int{{{\cos }^{5}}xdx}$

Ans: - Rewrite the given expression as 

$\int{{{\cos }^{5}}xdx}=\int{{{\cos }^{4}}x}\cos xdx$

$\Rightarrow \int{{{\cos }^{5}}xdx}=\int{{{\left( {{\cos }^{2}}x \right)}^{2}}\cos xdx}$

Use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the above equation.

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

Assume the substitution $u=\sin x\Leftrightarrow du=\cos xdx$ in the above equation.

$\int{{{\cos }^{5}}xdx}=\int{{{\left( 1-{{u}^{2}} \right)}^{2}}du}$

$\Rightarrow \int{{{\cos }^{5}}xdx}=\int{\left( 1-2{{u}^{2}}+{{u}^{4}} \right)du}$

$\Rightarrow \int{{{\cos }^{5}}xdx}=u-\frac{2}{3}{{u}^{3}}+\frac{1}{5}{{u}^{5}}+C$

Resubstitute the value $u=\sin x$ in the above equation.

$\Rightarrow \int{{{\cos }^{5}}xdx}=\sin x-\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+C$

Hence the value of $\int{{{\cos }^{5}}xdx}$ is $\sin x-\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+C$.


(vii). $\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}$

Ans: - Rewrite the given expression as 

$\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\int{\frac{1}{8}\left[ \left( 4{{\sin }^{2}}x{{\cos }^{2}}x \right)\left( 2{{\cos }^{2}}x \right) \right]dx}$

$\Rightarrow \int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{8}\int{\left[ {{\left( 2\sin x\cos x \right)}^{2}}\left( 2{{\cos }^{2}}x \right) \right]}dx$

Use the trigonometric formulas $2\sin A\cos A=\sin 2A$ and $\cos 2A=2{{\cos }^{2}}A-1$ in the above equation.

$\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{8}\int{{{\left( \sin 2x \right)}^{2}}\left( 1+\cos 2x \right)dx}$

$\Rightarrow \int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{8}\int{{{\sin }^{2}}2x\left( 1+\cos 2x \right)dx}$

Rewrite the above equation as

$\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{16}\int{\left( 2{{\sin }^{2}}2x \right)\left( 1+\cos 2x \right)dx}$

Use the trigonometric formula $\cos 2x=1-2{{\sin }^{2}}x$ in the above equation.

$\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{16}\int{\left( 1-\cos 4x \right)\left( 1+\cos 2x \right)dx}$

$\Rightarrow \int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{16}\int{\left[ 1-\cos 4x+\cos 2x-\cos 4x\cos 2x \right]dx}$

Use the trigonometric formula $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ in the above equation.

$\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{16}\int{\left[ 1-\cos 4x+\cos 2x-\frac{1}{2}\left( \cos 6x+\cos 2x \right) \right]}dx$

$\Rightarrow \int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{32}\int{\left( 2+\cos 2x-2\cos 4x-\cos 6x \right)dx}$

$\Rightarrow \int{{{\sin }^{2}}x{{\cos }^{4}}xdx}=\frac{1}{32}\left[ 2x+\frac{\sin 2x}{2}-\frac{2\sin 4x}{4}-\frac{\sin 6x}{6} \right]+C$

Hence the value of $\int{{{\sin }^{2}}x{{\cos }^{4}}xdx}$ is $\frac{1}{32}\left( 2x+\frac{1}{2}\sin 2x-\frac{1}{2}\sin 4x-\frac{1}{6}\sin 6x \right)+C$.


(viii). $\int{{{\cot }^{3}}x{{\csc }^{4}}xdx}$

Ans: - Rewrite the given expression by using trigonometric identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$.

$\int{{{\cot }^{3}}x{{\csc }^{4}}xdx}=\int{-\left( -{{\cot }^{2}}x\left( {{\cot }^{2}}x+1 \right) \right){{\csc }^{2}}xdx}$

Assume the substitution $u=\cot x\Leftrightarrow dx=-\frac{1}{{{\csc }^{2}}x}du$ in the above equation.

$\int{{{\cot }^{3}}x{{\csc }^{4}}xdx}=-\int{{{u}^{3}}\left( {{u}^{2}}+1 \right)du}$

$\Rightarrow \int{{{\cot }^{3}}x{{\csc }^{4}}xdx}=-\int{\left( {{u}^{5}}+{{u}^{3}} \right)du}$

$\Rightarrow \int{{{\cot }^{3}}x{{\csc }^{4}}xdx}=-\left[ \frac{{{u}^{6}}}{6}+\frac{{{u}^{4}}}{4} \right]+C$

Resubstitute the value $u=\cot x$ in the above equation.

$\int{{{\cot }^{3}}x{{\csc }^{4}}xdx}=-\left[ \frac{{{\cot }^{6}}x}{6}+\frac{{{\cot }^{4}}x}{4} \right]+C$

Hence the value of $\int{{{\cot }^{3}}x{{\csc }^{4}}xdx}$ is $-\left[ \frac{{{\cot }^{6}}x}{6}+\frac{{{\cot }^{4}}x}{4} \right]+C$.


(ix). $\int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}$

Ans: - Assume the substitution $u={{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x\Leftrightarrow dx=\frac{1}{2{{a}^{2}}\cos x\sin x-2{{b}^{2}}\cos x\sin x}du$ in the above equation.

$\int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}=\int{\frac{1}{2\left( {{a}^{2}}-{{b}^{2}} \right)\sqrt{u}}du}$

$\Rightarrow \int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}=\frac{1}{2\left( {{a}^{2}}-{{b}^{2}} \right)}\int{\frac{1}{\sqrt{u}}du}$

$\Rightarrow \int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}=\frac{1}{2\left( {{a}^{2}}-{{b}^{2}} \right)}\times 2\sqrt{u}+C$

Resubstitute the value $u={{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x$ in the above equation.

$\int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}=\frac{1}{{{a}^{2}}-{{b}^{2}}}\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}+C$

Hence the value of $\int{\frac{\sin x\cos x}{\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}}dx}$ is $\frac{1}{{{a}^{2}}-{{b}^{2}}}\sqrt{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}+C$.


(x). $\int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}$

Ans: - Use the trigonometric formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ in the above expression.

\[\int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=\int{\frac{1}{\sqrt{{{\cos }^{3}}x\left( \cos x\cos a-\sin x\sin a \right)}}dx}\]

$\Rightarrow \int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=\int{\frac{1}{\sqrt{{{\cos }^{4}}x\left( \cos a-\tan x\sin a \right)}}dx}$

$\Rightarrow \int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=\int{\frac{1}{{{\cos }^{2}}x\sqrt{\cos a-\tan x\sin a}}dx}$

$\Rightarrow \int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=\int{\frac{{{\sec }^{2}}x}{\sqrt{\cos a-\tan x\sin a}}dx}$

Assume the substitution $u=\cos a-\sin a\tan x\Leftrightarrow du=-\sin a{{\sec }^{2}}xdx$ since the terms $\sin a$, $\cos a$ are constants.

$\int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=-\int{\frac{1}{\sin a\sqrt{u}}du}$

$\Rightarrow \int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=-2\csc a\sqrt{u}+C$

Resubstitute the value $u=\cos a-\sin a\tan x$ in the above equation.

$\int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}=-2\csc a\sqrt{\cos a-\sin a\tan x}+C$

Hence the value of $\int{\frac{1}{\sqrt{{{\cos }^{3}}x\cos \left( x+a \right)}}dx}$ is $-2\csc a\sqrt{\cos a-\sin a\tan x}+C$.


(xi). $\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

Ans: - Modify the given expression as 

$\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\frac{{{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx$

Use the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ in the above equation.

$\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\frac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right)}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx$

$\Rightarrow \int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\frac{{{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}+2{{\sin }^{2}}x{{\cos }^{2}}x-2{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

\[\Rightarrow \int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\frac{{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-3{{\sin }^{2}}x{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}dx\]

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

$\Rightarrow \int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{{{\sec }^{2}}xdx}+\int{{{\csc }^{2}}x}dx-\int{3dx}+C$

$\Rightarrow \int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\tan x-\cot x-3x+C$

Hence the value of $\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ is $\tan x-\cot x-3x+C$.


(xii). $\int{\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$

Ans: - Assume the substitution $\sin x-\cos x=t\Leftrightarrow \left( \sin x+\cos x \right)dx=dt$ and the value of ${{t}^{2}}$ will be 

${{t}^{2}}={{\left( \sin x-\cos x \right)}^{2}}$

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

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

Substitute all the values in the given expression.

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

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

Substitute the value $t=\sin x-\cos x$ in the above equation.

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

Hence the value of $\int{\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$ is ${{\sin }^{-1}}\left( \sin x-\cos x \right)+C$.


42. Evaluate

(i). $\int{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}$

Ans: - Assume the substitution ${{x}^{2}}=t\Leftrightarrow 2xdx=dt$in the above equation.

$\int{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}=\int{\frac{1}{2\left( {{t}^{2}}+t+1 \right)}dt}$

$\Rightarrow \int{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}=\frac{1}{2}\int{\frac{1}{{{\left( t+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}dt$

$\Rightarrow \int{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}=\frac{1}{2}\times \frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)+C$

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

Substitute $t={{x}^{2}}$ in the above equation.

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

Hence the value of $\int{\frac{x}{{{x}^{4}}+{{x}^{2}}+1}dx}$ is $\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2{{x}^{2}}+1}{\sqrt{3}} \right)+C$.


(ii). $\int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}$

Ans: - Assume the substitution $t=\log x\Leftrightarrow dt=\frac{1}{x}dx$ in the above equation.

$\int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\int{\frac{dt}{6{{t}^{2}}+7t+2}}$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\int{\frac{dt}{6{{t}^{2}}+3t+4t+2}}$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\int{\frac{1}{\left( 3t+2 \right)\left( 2t+1 \right)}dt}$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\int{\frac{2\left( 3t+2 \right)-3\left( 2t+1 \right)}{\left( 3t+2 \right)\left( 2t+1 \right)}dt}$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\int{\frac{2dt}{2t+1}}-\int{\frac{3dt}{3t+2}}$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\log \left( 2t+1 \right)-\log \left( 3t+2 \right)+C$

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\log \left| \frac{2t+1}{3t+2} \right|+C$

Substitute the assumed value $t=\log x$ in the above equation.

$\Rightarrow \int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}=\log \left| \frac{2\log x+1}{3\log x+2} \right|+C$

Hence the value of $\int{\frac{1}{x\left[ 6{{\left( \log x \right)}^{2}}+7\log x+2 \right]}dx}$ is $\log \left| \frac{2\log x+1}{3\log x+2} \right|+C$.


(iii). $\int{\frac{dx}{1+x-{{x}^{2}}}}$

Ans: - Simplify the given expression as 

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

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

Assume the substitution $u=x-\frac{1}{2}\Leftrightarrow du=dx$ in the above equation.

$\int{\frac{dx}{1+x-{{x}^{2}}}}=\int{\frac{4}{-4{{u}^{2}}+5}}du$

Assume the substitution $u=\frac{\sqrt{5}}{3}v\Leftrightarrow du=dv$ in the above equation.

$\int{\frac{dx}{1+x-{{x}^{2}}}}=4\int{\frac{1}{2\sqrt{5}\left( -{{v}^{2}}+1 \right)}dv}$

$\Rightarrow \int{\frac{dx}{1+x-{{x}^{2}}}}=4\times \frac{1}{2\sqrt{5}}\int{\frac{1}{-{{v}^{2}}+1}dv}$

$\Rightarrow \int{\frac{dx}{1+x-{{x}^{2}}}}=\frac{2}{\sqrt{5}}\times \frac{1}{2}\left( \log \left| v+1 \right|-\log \left| v-1 \right| \right)+C$

Resubstitute $v=\frac{2}{\sqrt{5}}u$ and $u=x-\frac{1}{2}$ in the above equation.

$\int{\frac{dx}{1+x-{{x}^{2}}}}=\frac{1}{\sqrt{5}}\left[ \log \left( \frac{2}{\sqrt{5}}\left( x-\frac{1}{2} \right)+1 \right)-\log \left( \frac{2}{\sqrt{5}}\left( x-\frac{1}{2} \right)-1 \right) \right]+C$

$\Rightarrow \int{\frac{dx}{1+x-{{x}^{2}}}}=\frac{1}{\sqrt{5}}\log \left| \frac{\sqrt{5}-1+2x}{\sqrt{5}+1-2x} \right|+C$

Hence the value of $\int{\frac{dx}{1+x-{{x}^{2}}}}$ is $\frac{1}{\sqrt{5}}\log \left| \frac{\sqrt{5}-1+2x}{\sqrt{5}+1-2x} \right|+C$.


(iv). $\int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}$

Ans: - Simplify the above equation as 

$\int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}=\int{\frac{1}{\sqrt{9+2\left( 4 \right)x-{{x}^{2}}+{{4}^{2}}-{{4}^{2}}}}dx}$

$\Rightarrow \int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}=\int{\frac{1}{\sqrt{25-{{\left( x-4 \right)}^{2}}}}dx}$

$\Rightarrow \int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}=\int{\frac{1}{\sqrt{{{5}^{2}}-{{\left( x-4 \right)}^{2}}}}dx}$

Apply the integration formula $\int{\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=}{{\sin }^{-1}}\left( \frac{x}{a} \right)+C$ in the above equation.

$\int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \frac{x-4}{5} \right)+C$

Hence the value of $\int{\frac{1}{\sqrt{9+8x-{{x}^{2}}}}dx}$ is ${{\sin }^{-1}}\left( \frac{x-4}{5} \right)+C$.


(v) $\int{\frac{\mathbf{1}}{\sqrt{\left( \mathbf{x-a} \right)\left( \mathbf{x-b} \right)}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\frac{dx}{\sqrt{{{x}^{2}}-bx-ax+ab}}}$

$=\int{\frac{dx}{\sqrt{{{x}^{2}}-x\left( a+b \right)+ab}}}$

$=\int{\frac{dx}{\sqrt{{{x}^{2}}-2x\left( \frac{a+b}{2} \right)+{{\left( \frac{a+b}{2} \right)}^{2}}-{{\left( \frac{a+b}{2} \right)}^{2}}+ab}}}$

$=\int{\frac{dx}{\sqrt{{{\left( x-\frac{a+b}{2} \right)}^{2}}-\left( \frac{{{a}^{2}}+2ab+{{b}^{2}}}{4}+ab \right)}}}$

$=\int{\frac{dx}{\sqrt{{{\left( x-\frac{a+b}{2} \right)}^{2}}+\frac{-{{a}^{2}}-{{b}^{2}}+2ab}{4}}}}$

$=\int{\frac{dx}{\sqrt{{{\left( x-\frac{a+b}{2} \right)}^{2}}-{{\left( \frac{a-b}{2} \right)}^{2}}}}}$

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

After simplifying we get,

$\int{\frac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| x-\frac{a+b}{2}+\sqrt{{{x}^{2}}-x\left( a+b \right)+ab} \right|+C$
Thus, we’ll have

$\int{\frac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| x-\frac{a+b}{2}+\sqrt{\left( x-a \right)\left( x+b \right)} \right|+C$, where $C$ is an integration constant.


(vi) $\int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}}$

Ans:  Multiply and divide the fraction with $\sin \left( x-\alpha  \right)$.

$\int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}dx}=\int{\frac{\sin \left( x-\alpha  \right)}{\sqrt{\sin \left( x+\alpha  \right)\sin \left( x-\alpha  \right)}}dx}$

$\Rightarrow \int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}}dx=\int{\frac{\sin x\cos \alpha -\cos x\sin \alpha }{\sqrt{{{\sin }^{2}}x-{{\sin }^{2}}\alpha }}dx}$

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

$\Rightarrow \int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}}dx=\cos \alpha \int{\frac{\sin xdx}{\sqrt{{{\cos }^{2}}\alpha -{{\cos }^{2}}x}}}-\sin \alpha \int{\frac{\cos x}{\sqrt{{{\sin }^{2}}x-{{\sin }^{2}}\alpha }}dx}$

Assume $u=\cos x\Leftrightarrow du=-\sin xdx$ and $v=\sin x\Leftrightarrow dv=\cos xdx$ in the above equation.

$\Rightarrow \int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}dx}=\cos \alpha \int{\frac{-du}{\sqrt{{{\cos }^{2}}-{{u}^{2}}}}}-\sin \alpha \int{\frac{dv}{\sqrt{{{v}^{2}}-{{\sin }^{2}}\alpha }}}$

\[\Rightarrow \int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}dx}=-\cos \alpha {{\sin }^{-1}}\left( \frac{u}{\cos \alpha } \right)-\sin \alpha \log \left| v+\sqrt{{{v}^{2}}-{{\sin }^{2}}\alpha } \right|\]

Substitute the values of $u$ and $v$ in the above equation.

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

Hence the value of $\int{\sqrt{\frac{\sin \left( x-\alpha  \right)}{\sin \left( x+\alpha  \right)}}}dx$ is $-\cos \alpha {{\sin }^{-1}}\left( \frac{\cos x}{\cos \alpha } \right)-\sin \alpha \log \left| \sin x+\sqrt{{{\sin }^{2}}x-{{\sin }^{2}}\alpha } \right|+C$.


(vii). $\int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}$

Ans: - Write $5x-2$ as $\frac{5}{6}\left( 6x+2 \right)-\frac{11}{3}$ and split.

$\int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}=\int{\left( \frac{5\left( 6x+2 \right)}{6\left( 3{{x}^{2}}+2x+1 \right)}-\frac{11}{3\left( 3{{x}^{2}}+2x+1 \right)} \right)dx}$

Assume the substitution $u=3{{x}^{2}}+2x+1\Leftrightarrow dx=\frac{1}{6x+2}du$ in the above equation.

$\int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}=\frac{5}{6}\int{\frac{1}{u}du}-\frac{11}{3}\int{\frac{1}{{{\left( \sqrt{3}x+\frac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \sqrt{\frac{2}{3}} \right)}^{2}}}dx}$

$\Rightarrow \int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}=\frac{5}{6}\ln \left| u \right|-\frac{11}{3}\times \frac{1}{\left( \sqrt{\frac{2}{3}} \right)}{{\tan }^{-1}}\left( \frac{\sqrt{3}x+\frac{1}{\sqrt{3}}}{\sqrt{\frac{2}{3}}} \right)+C$

$\Rightarrow \int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}=\frac{5}{6}\log \left| 3{{x}^{2}}+2x+1 \right|-\frac{11}{3\sqrt{2}}{{\tan }^{-1}}\left( \frac{3x+1}{\sqrt{2}} \right)+C$

Hence the value of $\int{\frac{5x-2}{3{{x}^{2}}+2x+1}dx}$ is $\frac{5}{6}\log \left| 3{{x}^{2}}+2x+1 \right|-\frac{11}{3\sqrt{2}}{{\tan }^{-1}}\left( \frac{3x+1}{\sqrt{2}} \right)+C$.


(viii). $\int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx$

Ans: - Perform long division method and simplify the above fraction as 

$\int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=\int{\left( 1+\frac{-6x-12}{{{x}^{2}}+6x+12} \right)dx}$

$\Rightarrow \int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=\int{du}+\int{\frac{-6x-12}{{{x}^{2}}+2\left( 3 \right)x+{{3}^{2}}-{{3}^{2}}+12}dx}$

$\Rightarrow \int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=x+\int{\frac{-6x-12}{{{\left( x+3 \right)}^{2}}+2}dx}$

Assume the substitution $u=x+3\Leftrightarrow du=dx$ in the above equation.

$\int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=x+\int{\frac{-6u+6}{{{u}^{2}}+3}du}$

$\Rightarrow \int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=x-3\int{\frac{2u}{{{u}^{2}}+3}+6\int{\frac{1}{{{u}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}du}}$

$\Rightarrow \int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=x-\log \left| {{u}^{2}}+3 \right|+6\times \frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{u}{\sqrt{3}} \right)+C$

$\Rightarrow \int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx=x-3\log \left| {{x}^{2}}+6x+12 \right|+2\sqrt{3}{{\tan }^{-1}}\left( \frac{x+3}{\sqrt{3}} \right)+C$

Hence the value of $\int{\frac{{{x}^{2}}}{{{x}^{2}}+6x+12}}dx$ is $x-3\log \left| {{x}^{2}}+6x+12 \right|+2\sqrt{3}{{\tan }^{-1}}\left( \frac{x+3}{\sqrt{3}} \right)+C$.


(ix). $\int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}$

Ans: - Simplify the given expression as 

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

$\Rightarrow \int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}=\int{\frac{x+2}{\sqrt{-{{\left( x-2 \right)}^{2}}+4}}dx}$

Assume the substitution $u=x-2\Leftrightarrow du=dx$ in the above equation.

$\int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}=\int{\frac{u+4}{\sqrt{-{{u}^{2}}+4}}du}$

$\Rightarrow \int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}=\int{\frac{u}{\sqrt{-{{u}^{2}}+4}}du}+4\int{\frac{1}{\sqrt{-{{u}^{2}}+{{2}^{2}}}}du}$

$\Rightarrow \int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}=-\sqrt{-{{u}^{2}}+4}+4{{\sin }^{-1}}\left( \frac{u}{2} \right)$

Substitute the value $u=x-2$ in the above equation.

$\int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}=-\sqrt{4x-{{x}^{2}}}+4{{\sin }^{-1}}\left( \frac{x-2}{2} \right)+C$

Hence the value of $\int{\frac{x+2}{\sqrt{4x-{{x}^{2}}}}dx}$ is $-\sqrt{4x-{{x}^{2}}}+4{{\sin }^{-1}}\left( \frac{x-2}{2} \right)+C$.


(x). $\int{x\sqrt{1+x-{{x}^{2}}}}dx$

Ans: - Write the term $x$ as $\frac{1}{2}-\frac{1}{2}\left( 1-2x \right)$ in the above expression.

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

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

Assume the substitution $u=-{{x}^{2}}+x+1\Leftrightarrow dx=\frac{1}{1-2x}du$ in the above equation.

$\int{x\sqrt{1+x-{{x}^{2}}}}dx=\frac{1}{2}\int{-\sqrt{u}}du+\frac{1}{2}\int{\sqrt{-{{x}^{2}}+2\left( \frac{1}{2} \right)x+{{\left( \frac{1}{2} \right)}^{2}}-{{\left( \frac{1}{2} \right)}^{2}}+1}}dx$

$\Rightarrow \int{x\sqrt{1+x-{{x}^{2}}}}dx=-\frac{1}{2}\times \frac{2{{u}^{\frac{3}{2}}}}{3}+\frac{1}{2}\sqrt{{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}-{{\left( x-\frac{1}{2} \right)}^{2}}}dx$

Apply the integration formula $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right)+C$ in the above equation.

$\int{x\sqrt{1+x-{{x}^{2}}}}dx=-\frac{1}{3}{{\left( -{{x}^{2}}+x+1 \right)}^{\frac{3}{2}}}+\frac{1}{2}\times \frac{\left( x-\frac{1}{2} \right)}{2}\sqrt{-{{x}^{2}}+x+1}+\frac{1}{2}\times \frac{\frac{5}{4}}{2}{{\sin }^{-1}}\left( \frac{x-\frac{1}{2}}{\frac{\sqrt{5}}{2}} \right)+C$

$\Rightarrow \int{x\sqrt{1+x-{{x}^{2}}}}dx=-\frac{1}{3}{{\left( 1+x-{{x}^{2}} \right)}^{\frac{3}{2}}}+\frac{1}{8}\left( 2x-1 \right)\sqrt{1+x-{{x}^{2}}}+\frac{5}{16}{{\sin }^{-1}}\left( \frac{2x-1}{\sqrt{5}} \right)+C$

Hence the value of $\int{x\sqrt{1+x-{{x}^{2}}}}dx$ is $-\frac{1}{3}{{\left( 1+x-{{x}^{2}} \right)}^{\frac{3}{2}}}+\frac{1}{8}\left( 2x-1 \right)\sqrt{1+x-{{x}^{2}}}+\frac{5}{16}{{\sin }^{-1}}\left( \frac{2x-1}{\sqrt{5}} \right)+C$.


(xi). $\int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx$

Ans: - Write the term $3x-2$ as $\frac{3}{2}\left( 2x+1 \right)-\frac{7}{2}$ in the above expression.

$\int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx=\int{\left( \frac{3\left( 2x+1 \right)\sqrt{{{x}^{2}}+x+1}}{2}-\frac{7\sqrt{{{x}^{2}}+x+1}}{2} \right)}dx$

$\Rightarrow \int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx=\frac{3}{2}\int{\left( 2x+1 \right)\sqrt{{{x}^{2}}+x+1}}dx-\frac{7}{2}\int{\sqrt{{{x}^{2}}+x+1}}dx$

Assume the substitution $u={{x}^{2}}+x+1\Leftrightarrow dx=\frac{1}{2x+1}du$ in the above equation.

$\int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx=\frac{3}{2}\times \int{\sqrt{u}du}-\frac{7}{2}\int{\sqrt{{{x}^{2}}+2\left( \frac{1}{2} \right)x+{{\left( \frac{1}{2} \right)}^{2}}-{{\left( \frac{1}{2} \right)}^{2}}+1}}dx$

$\Rightarrow \int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx=\frac{3}{2}\times \frac{2}{3}{{u}^{\frac{3}{2}}}-\frac{7}{2}\int{\sqrt{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}dx$

Apply the integration formula $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx}=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C$ in the above equation.

$\int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx={{\left( {{x}^{2}}+x+1 \right)}^{\frac{3}{2}}}-\frac{7}{2}\left[ \frac{x+\frac{1}{2}}{2}\sqrt{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}+\frac{\frac{3}{4}}{2}\log \left| x+\frac{1}{2}+\sqrt{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}} \right| \right]+C$

$\Rightarrow \int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx={{\left( {{x}^{2}}+x+1 \right)}^{\frac{3}{2}}}-\frac{7}{2}\left[ \left( x+\frac{1}{2} \right)\sqrt{{{x}^{2}}+x+1}+\frac{3}{8}\log \left| x+\frac{1}{2}+\sqrt{{{x}^{2}}+x+1} \right| \right]+C$

Hence the value of $\int{\left( 3x-2 \right)\sqrt{{{x}^{2}}+x+1}}dx$ is ${{\left( {{x}^{2}}+x+1 \right)}^{\frac{3}{2}}}-\frac{7}{2}\left[ \left( x+\frac{1}{2} \right)\sqrt{{{x}^{2}}+x+1}+\frac{3}{8}\log \left| x+\frac{1}{2}+\sqrt{{{x}^{2}}+x+1} \right| \right]+C$.


(xii). $\int{\sqrt{\sec x+1}}dx$

Ans: We know that,

$\text{secx=}\frac{\text{1}}{\text{cosx}}$

So, the given integral can be written as 

$\int{\sqrt{\sec x+1}\text{ }dx}=\int{\sqrt{\frac{1}{\cos x}+1}dx}$

$=\int{\sqrt{\frac{1+\cos x}{\cos x}}dx}$

Since, we have the half angle formulas as,

$1+\cos 2x=2{{\cos }^{2}}\frac{x}{2}$

$\cos 2x=1-2{{\sin }^{2}}\frac{x}{2}$

Using these formulas in the integral we get,

$=\int{\sqrt{\frac{2{{\cos }^{2}}\frac{x}{2}}{1-2{{\sin }^{2}}\frac{x}{2}}}dx}$

$=\sqrt{2}\int{\frac{\cos \frac{x}{2}}{\sqrt{1-2{{\sin }^{2}}\frac{x}{2}}}dx}$

Taking $\sqrt{2}$ out from the denominator we get,

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

Now, let us assume,

$\sin \frac{\text{x}}{\text{2}}=\text{k}$

Now, differentiating on both sides we have,

\[\begin{align}

  & \frac{1}{2}\cos \frac{\text{x}}{2}\text{dx}=\text{dk} \\ 

 & \cos \frac{\text{x}}{2}\text{dx}=2\text{dk} \\ 

\end{align}\]

Now the integral becomes,

\[=\int{\frac{2dk}{\sqrt{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{\left( k \right)}^{2}}}}}\]

We know the formula,

$\int{\frac{\text{dx}}{\sqrt{{{\text{a}}^{2}}-{{\text{x}}^{2}}}}}={{\sin }^{-1}}\left( \frac{\text{x}}{\text{a}} \right)+c$

So, we get the integral value as,

$=2{{\sin }^{-1}}\left( \frac{\text{k}}{\left( \frac{1}{\sqrt{2}} \right)} \right)+c$

\[=2{{\sin }^{-1}}\left( \frac{\sin \frac{x}{2}}{\frac{1}{\sqrt{2}}} \right)+c\]

Hence, $\int{\sqrt{\sec x+1}\text{ }dx}=2{{\sin }^{-1}}\left( \sqrt{2}\sin \frac{x}{2} \right)+c$, where $c$ is an integration constant.


43. Evaluate: 

(i). $\int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}$

Ans: - Assume the substitution $u={{x}^{7}}\Leftrightarrow du=7{{x}^{6}}dx$ in the above equation.

$\int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}=\int{\frac{1}{x\left( {{x}^{7}}+1 \right)}}\times \frac{1}{7{{x}^{6}}}du$

$\Rightarrow \int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}=\frac{1}{7}\int{\frac{1}{u\left( u+1 \right)}du}$

Perform the partial fraction decomposition in the above equation.

$\int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}=\frac{1}{7}\int{\left( \frac{1}{u}-\frac{1}{u+1} \right)du}$

$\Rightarrow \int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}=\frac{1}{7}\left[ \log \left| u \right|-\log \left| u+1 \right| \right]+C$

$\Rightarrow \int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}=\frac{1}{7}\log \left| \frac{{{x}^{7}}}{{{x}^{7}}+1} \right|+C$

Hence the value of $\int{\frac{dx}{x\left( {{x}^{7}}+1 \right)}}$ is $\frac{1}{7}\log \left| \frac{{{x}^{7}}}{{{x}^{7}}+1} \right|+C$.


(ii). $\int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}$

Ans: - Assume the substitution $\cos x=t\Leftrightarrow \sin xdx=-dt$ in the above equation.

$\int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}=\int{\frac{-dt}{\left( 1+t \right)\left( 2+3t \right)}}$

On performing the partial fraction decomposition, the above fraction is modified as 

$\int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}=\int{\left( \frac{1}{1+t}-\frac{3}{2+3t} \right)dt}$

$\Rightarrow \int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}=\log \left| 1+t \right|-\log \left| 2+3t \right|+C$

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

$\Rightarrow \int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}=\log \left| \frac{1+\cos x}{2+3\cos x} \right|+C$

Hence the value of $\int{\frac{\sin x}{\left( 1+\cos x \right)\left( 2+3\cos x \right)}dx}$ is $\log \left| \frac{1+\cos x}{2+3\cos x} \right|+C$.


(iii). $\int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }$

Ans: - Assume the substitution $u=\cos \theta \Leftrightarrow du=-\sin \theta d\theta $ in the above equation.

$\int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }=\int{-\frac{u}{{{u}^{2}}-u-2}du}$

Perform partial fraction decomposition in the above equation.

$\int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }=-\int{\left( \frac{1}{3\left( u+1 \right)}+\frac{2}{3\left( u-2 \right)} \right)}dx$

$\Rightarrow \int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }=-\frac{1}{3}\int{\frac{1}{u+1}du}-\frac{2}{3}\int{\frac{1}{u-2}du}$

$\Rightarrow \int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }=-\frac{1}{3}\log \left| u+1 \right|-\frac{2}{3}\log \left| u-2 \right|+C$

Substitute the value $u=\cos \theta $ in the above equation.

$\int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }=-\frac{2}{3}\log \left| \cos \theta -2 \right|-\frac{1}{3}\log \left| 1+\cos \theta  \right|+C$.

Hence the value of $\int{\frac{\sin \theta \cos \theta }{{{\cos }^{2}}\theta -\cos \theta -2}d\theta }$ is $-\frac{2}{3}\log \left| \cos \theta -2 \right|-\frac{1}{3}\log \left| 1+\cos \theta  \right|+C$.


(iv). $\int{\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}dx}$

Ans: Let us represent the above expression in the partial fractions as,

$\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+3}$

Taking the LCM in above equation and representing as the coefficients of $\text{x}$ we get,

$x-1={{x}^{2}}\left( A+B+C \right)+x\left( A+4B-C \right)+\left( -6A+3B-2C \right)$

Comparing the coefficients on both sides of equation we get,

$\begin{align} & A+B+C=0 \\ & A+4B-C=1 \\ & -6A+3B-2C=-1 \\ \end{align}$

On solving the above three equations we get,

$\begin{align} & A=\frac{1}{3} \\ & B=\frac{1}{15} \\ & C=\frac{-2}{5} \\ \end{align}$

So, the integral becomes,

$\int{\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}dx}=\int{\left( \frac{1}{3\left( x+1 \right)}+\frac{1}{15\left( x-2 \right)}-\frac{2}{5\left( x+3 \right)} \right)dx}$

$\Rightarrow \int{\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}dx}=\frac{1}{3}\int{\frac{1}{x+1}dx}+\frac{1}{15}\int{\frac{1}{x-2}dx}-\frac{2}{5}\int{\frac{1}{x+3}dx}$

$\Rightarrow \int{\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}dx}=\frac{1}{3}\log \left| x+1 \right|+\frac{1}{15}\log \left| x-2 \right|-\frac{2}{5}\log \left| x+3 \right|+C$

Hence the value of  the integral $\int{\frac{x-1}{\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)}dx}$ is $\frac{1}{3}\log \left| x+1 \right|+\frac{1}{15}\log \left| x-2 \right|-\frac{2}{5}\log \left| x+3 \right|+C$.


(v). $\int{\frac{{{x}^{2}}+x+2}{\left( x-2 \right)\left( x-1 \right)}dx}$

Ans: - Simplify the given expression by performing long division.

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

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

Perform partial fraction decomposition in the above equation.

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

$\Rightarrow \int{\frac{{{x}^{2}}+x+2}{\left( x-2 \right)\left( x-1 \right)}dx}=4\left[ 2\int{\frac{1}{x-2}dx-\int{\frac{1}{x-1}dx}} \right]+x$

$\Rightarrow \int{\frac{{{x}^{2}}+x+2}{\left( x-2 \right)\left( x-1 \right)}dx}=8\log \left| x-2 \right|-4\log \left| x-1 \right|+x+C$

$\Rightarrow \int{\frac{{{x}^{2}}+x+2}{\left( x-2 \right)\left( x-1 \right)}dx}=x+4\log \left| \frac{{{\left( x-2 \right)}^{2}}}{x-1} \right|+C$

Hence the value of $\int{\frac{{{x}^{2}}+x+2}{\left( x-2 \right)\left( x-1 \right)}dx}$ is $x+4\log \left| \frac{{{\left( x-2 \right)}^{2}}}{x-1} \right|+C$.


(vi). $\int{\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}dx}$

Ans: - Assume the substitution $t={{x}^{2}}$ in the given fraction.

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

$\Rightarrow \frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\frac{{{t}^{2}}+3t+2}{{{t}^{2}}+7t+12}$

Perform log division on the above fraction, then we will get 

$\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=1+\frac{-4t-10}{{{t}^{2}}+7t+12}$

Perform the partial fraction decomposition to the above fraction, then we will have 

$\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=1-\frac{2}{{{x}^{2}}+3}+\frac{6}{{{x}^{2}}+4}$

Apply integration on both sides of the above equation.

$\int{\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}dx}=\int{\left( 1-\frac{2}{{{x}^{2}}+3}+\frac{6}{{{x}^{2}}+4} \right)dx}$

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

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

Hence the value of $\int{\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}dx}$ is $x+\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{x}{\sqrt{3}} \right)-3{{\tan }^{-1}}\left( \frac{x}{2} \right)+C$.


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

Ans: - On performing partial fraction decomposition to the given expression we will have

$\int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}=\int{\left( \frac{4}{17\left( 2x+1 \right)}-\frac{2x-1}{17\left( {{x}^{2}}+4 \right)} \right)}dx$

$\Rightarrow \int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{4}{17}\int{\frac{1}{2x+1}dx}-\frac{1}{17}\int{\frac{2x-1}{{{x}^{2}}+4}dx}$

$\Rightarrow \int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{4}{17}\times \frac{1}{2}\log \left| 2x+1 \right|-\frac{1}{17}\left[ \int{\frac{2x}{{{x}^{2}}+4}dx-\int{\frac{1}{{{x}^{2}}+{{2}^{2}}}}}dx \right]$

$\Rightarrow \int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{2}{17}\log \left| 2x+1 \right|-\frac{1}{17}\left[ \log \left| {{x}^{2}}+4 \right|-\frac{1}{2}{{\tan }^{-1}}\left( \frac{x}{2} \right) \right]+C$

$\Rightarrow \int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{2}{17}\log \left| 2x+1 \right|-\frac{1}{17}\log \left| {{x}^{2}}+4 \right|+\frac{1}{34}{{\tan }^{-1}}\left( \frac{x}{2} \right)+C$

Hence the value of $\int{\frac{dx}{\left( 2x+1 \right)\left( {{x}^{2}}+4 \right)}}$ is $\frac{2}{17}\log \left| 2x+1 \right|-\frac{1}{17}\log \left| {{x}^{2}}+4 \right|+\frac{1}{34}{{\tan }^{-1}}\left( \frac{x}{2} \right)+C$.


(viii). $\int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}$

Ans: - Simplify the given fraction as 

$\int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}=-\int{\frac{dx}{\left( 2\cos x-1 \right)\sin x}\times \frac{\sin x}{\sin x}}$

Use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the above equation.

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

Assume the substitution $u=\cos x\Leftrightarrow du=-\sin xdx$ in the above equation.

$\int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}=\int{\frac{1}{\left( 2u-1 \right)\left( {{u}^{2}}-{{1}^{2}} \right)}du}$

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

Perform partial fraction decomposition in the above equation.

$\int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}=\int{\left( -\frac{4}{3\left( 2u-1 \right)}+\frac{1}{6\left( u+1 \right)}+\frac{1}{2\left( u-1 \right)} \right)}du$

$\Rightarrow \int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}=-\frac{4}{3}\times \frac{1}{2}\log \left| 2u-1 \right|+\frac{1}{6}\log \left| u+1 \right|+\frac{1}{2}\log \left| u-1 \right|+C$

$\Rightarrow \int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}=-\frac{1}{2}\log \left| 1-\cos x \right|-\frac{1}{6}\log \left| 1+\cos x \right|+\frac{2}{3}\log \left| 1-2\cos x \right|+C$

Hence the value of $\int{\frac{dx}{\sin x\left( 1-2\cos x \right)}}$ is $-\frac{1}{2}\log \left| 1-\cos x \right|-\frac{1}{6}\log \left| 1+\cos x \right|+\frac{2}{3}\log \left| 1-2\cos x \right|+C$.


(ix). $\int{\frac{\sin x}{\sin 4x}dx}$

Ans: - Use the trigonometric formulas $\sin 2A=2\sin A\cos A$ in the above equation.

$\int{\frac{\sin x}{\sin 4x}dx}=\int{\frac{\sin x}{2\sin 2x\cos 2x}dx}$

$\Rightarrow \int{\frac{\sin x}{\sin 4x}dx}=\int{\frac{\sin x}{2\left( 2\sin x\cos x \right)\cos 2x}\times \frac{\cos x}{\cos x}dx}$

$\Rightarrow \int{\frac{\sin x}{\sin 4x}dx}=\frac{1}{4}\int{\frac{\cos x}{{{\cos }^{2}}x\cos 2x}dx}$

Use the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\cos 2A=1-2{{\sin }^{2}}A$ in the above equation.

$\int{\frac{\sin x}{\sin 4x}dx}=\frac{1}{4}\int{\frac{\cos xdx}{\left( 1-{{\sin }^{2}}x \right)\left( 1-2{{\sin }^{2}}x \right)}}$

Assume the substitution $\sin x=t\Leftrightarrow \cos xdx=dt$ in the above equation.

$\int{\frac{\sin x}{\sin 4x}dx}=\frac{1}{4}\int{\frac{1}{\left( 1-{{t}^{2}} \right)\left( 1-2{{t}^{2}} \right)}dt}$

Perform partial fraction decomposition.

$\int{\frac{\sin x}{\sin 4x}dx}=\frac{1}{4}\int{\left[ -\frac{1}{1-{{t}^{2}}}+\frac{2}{1-2{{t}^{2}}} \right]dt}$

$\Rightarrow \int{\frac{\sin x}{\sin 4x}dx}=-\frac{1}{4}\int{\frac{1}{1-{{t}^{2}}}dt}+\frac{2}{4}\int{\frac{1}{1-{{\left( \sqrt{2}t \right)}^{2}}}dt}$ 

$\Rightarrow \int{\frac{\sin x}{\sin 4x}dx}=-\frac{1}{4}\times \frac{1}{2}\log \left| \frac{1+t}{1-t} \right|+\frac{1}{2}\times \frac{1}{2\sqrt{2}}\log \left| \frac{1+\sqrt{2}t}{1-\sqrt{2}t} \right|+C$

$\Rightarrow \int{\frac{\sin x}{\sin 4x}dx}=-\frac{1}{8}\log \left| \frac{1+\sin x}{1-\sin x} \right|+\frac{1}{4\sqrt{2}}\log \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right|+C$

Hence the value of $\int{\frac{\sin x}{\sin 4x}dx}$ is $-\frac{1}{8}\log \left| \frac{1+\sin x}{1-\sin x} \right|+\frac{1}{4\sqrt{2}}\log \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right|+C$.


(x). $\int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}$

Ans: - Simplify the above expression as 

$\int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}=\int{\frac{1-\frac{1}{{{x}^{2}}}}{{{x}^{2}}+1+\frac{1}{{{x}^{2}}}}dx}$

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

Assume the substitution $t=x+\frac{1}{x}\Leftrightarrow dt=\left( 1-\frac{1}{{{x}^{2}}} \right)dx$ in the above equation.

$\int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}=\int{\frac{dt}{{{t}^{2}}-1}}$

$\Rightarrow \int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}=\frac{1}{2}\int{\frac{2dt}{\left( t+1 \right)\left( t-1 \right)}}$

Perform partial fraction decomposition in the above equation.

$\int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}=\frac{1}{2}\int{\frac{dt}{t-1}-\frac{1}{2}\int{\frac{dt}{t+1}}}$

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

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

$\Rightarrow \int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}=\frac{1}{2}\log \left| \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right|+C$

Hence the value of $\int{\frac{{{x}^{2}}-1}{{{x}^{4}}+{{x}^{2}}+1}dx}$ is $\frac{1}{2}\log \left| \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right|+C$.


(xi). $\int{\sqrt{\tan x}}dx$

Ans: - Assume the substitution $\tan x={{t}^{2}}\Leftrightarrow dx=\frac{2tdt}{{{\sec }^{2}}x}=\frac{2tdt}{1+{{t}^{4}}}$ in the given expression.

$\int{\sqrt{\tan x}}dx=\int{\left( t\times \frac{2t}{1+{{t}^{4}}} \right)dt}$

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

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

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

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

Assume the substitutions $t-\frac{1}{t}=u\Leftrightarrow \left( 1+\frac{1}{{{t}^{2}}} \right)dt=du$ and $t+\frac{1}{t}=v\Leftrightarrow \left( 1-\frac{1}{{{t}^{2}}} \right)dt=dv$ in the above equation.

$\int{\sqrt{\tan x}}dx=\int{\frac{1}{{{\left( t-\frac{1}{t} \right)}^{2}}+2}}du+\int{\frac{1}{{{\left( t+\frac{1}{t} \right)}^{2}}-2}dv}$

$\Rightarrow \int{\sqrt{\tan x}}dx=\int{\frac{1}{{{u}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}du}+\int{\frac{1}{{{v}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}}dv$

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

Substitute the values of $u$, $v$ and $t$ in the above equation.

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

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

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

Hence the value of $\int{\sqrt{\tan x}}dx$ is $\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\tan x-1}{\sqrt{2\tan x}} \right)+\frac{1}{2\sqrt{2}}\log \left| \frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1} \right|+C$.


(xii). $\int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}$

Ans: - Simplify the given expression by taking ${{x}^{2}}$ as common from the both numerator and denominator.

$\int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\int{\frac{1+\frac{9}{{{x}^{2}}}}{{{x}^{2}}+\frac{81}{{{x}^{2}}}}}dx$

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\int{\frac{1+\frac{9}{{{x}^{2}}}}{{{x}^{2}}+{{\left( \frac{9}{x} \right)}^{2}}+2\left( x \right)\left( \frac{9}{x} \right)-2\left( x \right)\left( \frac{9}{x} \right)}}dx$

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\int{\frac{1+\frac{9}{{{x}^{2}}}}{{{\left( x-\frac{9}{x} \right)}^{2}}+18}dx}$

Assume the substitution $x-\frac{9}{x}=t\Leftrightarrow \left( 1+\frac{9}{{{x}^{2}}} \right)dx=dt$ in the above equation.

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\int{\frac{dt}{{{t}^{2}}+{{\left( 3\sqrt{2} \right)}^{2}}}}$

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\frac{1}{3\sqrt{2}}{{\tan }^{-1}}\left( \frac{t}{3\sqrt{2}} \right)+C$

Substitute the value $t=x-\frac{9}{x}$ in the above equation.

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\frac{1}{3\sqrt{2}}{{\tan }^{-1}}\left( \frac{x-\frac{9}{x}}{3\sqrt{2}} \right)+C$

$\Rightarrow \int{\frac{{{x}^{2}}+9}{{{x}^{4}}+81}}=\frac{1}{3\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-9}{3\sqrt{2}} \right)+C$


44.

(i). $\int{{{x}^{5}}\sin {{x}^{3}}dx}$

Ans: - Simplify the above expression as 

$\int{{{x}^{5}}\sin {{x}^{3}}dx}=\int{{{x}^{3}}\sin \left( {{x}^{3}} \right)}{{x}^{2}}dx$

Assume the substitution ${{x}^{3}}=t\Leftrightarrow 3{{x}^{2}}dx=dt$ in the above equation.

$\int{{{x}^{5}}\sin {{x}^{3}}dx}=\int{t\sin t\times \frac{dt}{3}}$

To use the integration by parts rule, assume the functions $u=t\Leftrightarrow du=dt$ and $dv=\sin tdt\Leftrightarrow v=-\cos t$. Now use the integration by parts formula in the above equation.

$\int{{{x}^{5}}\sin {{x}^{3}}dx}=\frac{1}{3}\left[ t\left( -\cos t \right)-\int{\left( -\cos t \right)dt} \right]$

$\Rightarrow \int{{{x}^{5}}\sin {{x}^{3}}dx}=\frac{1}{3}\left[ -t\cos t+\int{\cos tdt} \right]$

$\Rightarrow \int{{{x}^{5}}\sin {{x}^{3}}dx}=\frac{1}{3}\left( -t\cos t+\sin t \right)+C$

Substitute the value $t={{x}^{3}}$ in the above equation.

$\int{{{x}^{5}}\sin {{x}^{3}}dx}=\frac{1}{3}\left( -{{x}^{3}}\cos \left( {{x}^{3}} \right)+\sin \left( {{x}^{3}} \right) \right)+C$

Hence the value of $\int{{{x}^{5}}\sin {{x}^{3}}dx}$ is $\frac{1}{3}\left( -{{x}^{3}}\cos \left( {{x}^{3}} \right)+\sin \left( {{x}^{3}} \right) \right)+C$.


(ii). $\int{{{\sec }^{3}}xdx}$

Ans: - Separate the function in the given expression as 

$\int{{{\sec }^{3}}xdx}=\int{\sec x\left( {{\sec }^{2}}x \right)dx}$

To use the integration by parts rule, assume the functions $u=\sec x\Leftrightarrow du=\sec x\tan xdx$ and $dv={{\sec }^{2}}xdx\Leftrightarrow v=\tan x$. Now substitute the values in the integration by parts formula.

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

Use the trigonometric identity ${{\sec }^{2}}x-\tan {{x}^{2}}=1$ in the above equation.

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

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

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

$\Rightarrow \int{{{\sec }^{3}}xdx}=\frac{1}{2}\left[ secx+\tan x+\ln \left| \sec x+\tan x \right| \right]+C$

Hence the value of $\int{{{\sec }^{3}}xdx}$ is $\frac{1}{2}\left[ secx+\tan x+\ln \left| \sec x+\tan x \right| \right]+C$.


(iii). $\int{{{e}^{ax}}\cos \left( bx+c \right)dx}$

Ans: - To use the integration by parts rule assume the functions $u={{e}^{ax}}\Leftrightarrow du=a{{e}^{ax}}$ and $dv=\cos \left( bx+c \right)\Leftrightarrow v=\frac{1}{b}\sin \left( bx+c \right)$. Now substitute the values in the integration by parts formula.

$\int{{{e}^{ax}}\cos \left( bx+c \right)dx}={{e}^{ax}}\left( \frac{1}{b}\sin \left( bx+c \right) \right)-\int{a{{e}^{ax}}\left( \frac{1}{b}\sin \left( bx+c \right) \right)dx}$

$\Rightarrow \int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{b}\sin \left( bx+c \right)-\frac{a}{b}\int{{{e}^{ax}}\sin \left( bx+c \right)dx}$

Assume the integral value $\int{{{e}^{ax}}\sin \left( bx+c \right)dx}$ separately from the above equation. For this value also assume the functions to use integration by parts as $u={{e}^{ax}}\Leftrightarrow du=a{{e}^{ax}}dx$ and $dv=\sin \left( bx+c \right)\Leftrightarrow v=-\frac{1}{b}\cos \left( bx+c \right)$. Substitute the values in the assumed value.

$\int{{{e}^{ax}}\sin \left( bx+c \right)dx}={{e}^{ax}}\left( -\frac{1}{b}\cos \left( bx+c \right) \right)-\int{a{{e}^{ax}}\left( -\frac{1}{b}\cos \left( bx+c \right) \right)}dx$

$\Rightarrow \int{{{e}^{ax}}\sin \left( bx+c \right)dx}=-\frac{{{e}^{ax}}}{b}\cos \left( bx+c \right)+\frac{a}{b}\int{{{e}^{ax}}\cos \left( bx+c \right)}dx$

Substitute the above value in the integral value of $\int{{{e}^{ax}}\cos \left( bx+c \right)dx}$.

$\int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{b}\sin \left( bx+c \right)-\frac{a}{b}\left[ -\frac{{{e}^{ax}}}{b}\cos \left( bx+c \right)+\frac{a}{b}\int{{{e}^{ax}}\cos \left( bx+c \right)}dx \right]$

$\Rightarrow \int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{b}\sin \left( bx+c \right)+\frac{a{{e}^{ax}}}{{{b}^{2}}}\cos \left( bx+c \right)-\frac{{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\cos \left( bx+c \right)}dx$

$\Rightarrow \int{{{e}^{ax}}\cos \left( bx+c \right)dx}+\frac{{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{{{b}^{2}}}\left[ b\sin \left( bx+c \right)+a\cos \left( bx+c \right) \right]+C$

$\Rightarrow \left( \frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \right)\int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{{{b}^{2}}}\left[ b\sin \left( bx+c \right)+a\cos \left( bx+c \right) \right]+C$

$\Rightarrow \int{{{e}^{ax}}\cos \left( bx+c \right)dx}=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ b\sin \left( bx+c \right)+a\cos \left( bx+c \right) \right]+C$

Hence the value of $\int{{{e}^{ax}}\cos \left( bx+c \right)dx}$ is $\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ b\sin \left( bx+c \right)+a\cos \left( bx+c \right) \right]+C$.


(iv). $\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}$

Ans: - Assume the substitution $3x=\tan t\Leftrightarrow 3dx={{\sec }^{2}}tdt$ in the above equation.

$\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\int{{{\sin }^{-1}}\left( \frac{2\tan t}{1+{{\tan }^{2}}t} \right)\times \frac{1}{3}{{\sec }^{2}}tdt}$

Use the formula $\sin 2x=\frac{2\tan x}{1+{{\tan }^{2}}x}$ in the above equation.

$\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\frac{1}{3}\int{{{\sin }^{-1}}\left( \sin 2t \right){{\sec }^{2}}tdt}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\frac{2}{3}\int{t{{\sec }^{2}}t}dt$

To use the integral by parts rule, assume the functions $u=t\Leftrightarrow du=dt$ and $dv={{\sec }^{2}}tdt\Leftrightarrow v=\tan t$ in the above equation.

$\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\frac{2}{3}\left[ t\left( \tan t \right)-\int{\tan tdt} \right]$

$\Rightarrow \int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\frac{2}{3}t\tan t-\frac{2}{3}\log \left| \sec t \right|+C$

Substitute the value $3x=\tan t\Leftrightarrow t={{\tan }^{-1}}\left( 3x \right)$ in the above equation.

$\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=\frac{2}{3}\left( {{\tan }^{-1}}\left( 3x \right) \right)\times 3x-\frac{1}{3}\log \left| 1+{{\tan }^{2}}t \right|+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}=2x{{\tan }^{-1}}3x-\frac{1}{3}\log \left| 1+9{{x}^{2}} \right|+C$

Hence the value of $\int{{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)dx}$ is $2x{{\tan }^{-1}}3x-\frac{1}{3}\log \left| 1+9{{x}^{2}} \right|+C$.


(v). $\int{\cos \sqrt{x}dx}$

Ans: - Assume the substitution $\sqrt{x}=t\Leftrightarrow \frac{1}{2\sqrt{x}}dx=dt$ in the given expression.

$\int{\cos \sqrt{x}dx}=\int{\left( \cos t \right)2tdt}$

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

To use the integration by parts rule, assume the functions $u=t\Leftrightarrow du=dt$ and $dv=\cos tdt\Leftrightarrow v=\sin t$.

$\Rightarrow \int{\cos \sqrt{x}dx}=2\left[ t\sin t-\int{\sin tdt} \right]$

$\Rightarrow \int{\cos \sqrt{x}dx}=2\left[ t\sin t-\left( -\cos t \right) \right]+C$

Substitute the value of $t=\sqrt{x}$ in the above equation.

$\int{\cos \sqrt{x}dx}=2\left[ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right]+C$

Hence the value of $\int{\cos \sqrt{x}dx}$ is $2\left[ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right]+C$.


(vi). $\int{{{x}^{3}}{{\tan }^{-1}}xdx}$

Ans: - To use the integration by parts rule assume the functions $u={{\tan }^{-1}}x\Leftrightarrow du=\frac{1}{1+{{x}^{2}}}dx$ and $dv={{x}^{3}}dx\Leftrightarrow v=\frac{{{x}^{4}}}{4}$.

$\int{{{x}^{3}}{{\tan }^{-1}}xdx}=\left( {{\tan }^{-1}}x \right)\left( \frac{{{x}^{4}}}{4} \right)-\int{\frac{{{x}^{4}}}{4}\left( \frac{1}{1+{{x}^{2}}} \right)dx}$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\frac{{{x}^{4}}{{\tan }^{-1}}x}{4}-\frac{1}{4}\int{\frac{\left( {{x}^{4}}-1 \right)+1}{1+{{x}^{2}}}dx}$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\frac{{{x}^{4}}{{\tan }^{-1}}x}{4}-\frac{1}{4}\int{\frac{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)+1}{1+{{x}^{2}}}dx}$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\frac{{{x}^{4}}{{\tan }^{-1}}x}{4}-\frac{1}{4}\left[ \int{\left( {{x}^{2}}-1 \right)dx+\int{\frac{1}{1+{{x}^{2}}}dx}} \right]$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\frac{{{x}^{4}}{{\tan }^{-1}}x}{4}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3}-x+{{\tan }^{-1}}x \right]+C$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\frac{{{x}^{4}}}{4}{{\tan }^{-1}}-\frac{{{x}^{3}}}{12}+\frac{x}{4}-\frac{1}{4}\tan x+C$

$\Rightarrow \int{{{x}^{3}}{{\tan }^{-1}}xdx}=\left( \frac{{{x}^{4}}-1}{4} \right){{\tan }^{-1}}x-\frac{{{x}^{4}}}{12}+\frac{x}{4}+C$

Hence the value of $\int{{{x}^{3}}{{\tan }^{-1}}xdx}$ is $\left( \frac{{{x}^{4}}-1}{4} \right){{\tan }^{-1}}x-\frac{{{x}^{4}}}{12}+\frac{x}{4}+C$.


(vii). $\int{{{e}^{2x}}\left( \frac{1+\sin 2x}{1+\cos 2x} \right)dx}$

Ans: - Use the trigonometric formulas $\sin 2x=2\sin x\cos x$ and $\cos 2x=2{{\cos }^{2}}x-1$ in the given expression.

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

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

$\Rightarrow \int{{{e}^{2x}}\left( \frac{1+\sin 2x}{1+\cos 2x} \right)dx}=\frac{1}{2}\int{{{e}^{2x}}{{\sec }^{2}}xdx+\int{{{e}^{2x}}\tan xdx}}$

Assume the functions $u={{e}^{2x}}\Leftrightarrow du=2{{e}^{2x}}dx$, $dv={{\sec }^{2}}xdx\Leftrightarrow v=\tan x$for the first integral in the above equation.

$\int{{{e}^{2x}}\left( \frac{1+\sin 2x}{1+\cos 2x} \right)dx}=\frac{1}{2}\left[ {{e}^{2x}}\tan x-\int{2{{e}^{2x}}\tan xdx} \right]+\int{{{e}^{2x}}\tan xdx}$

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

$\Rightarrow \int{{{e}^{2x}}\tan xdx}=\frac{1}{2}{{e}^{2x}}\tan x+C$

Hence the value of $\int{{{e}^{2x}}\tan xdx}$ is $\frac{1}{2}{{e}^{2x}}\tan x+C$.


(viii). $\int{{{e}^{x}}\left( \frac{x-1}{2{{x}^{2}}} \right)dx}$

Ans: - Simplify the given expression as 

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

Assume $t=\frac{1}{x}\Leftrightarrow {{t}^{'}}=-\frac{1}{{{x}^{2}}}$ in the above equation.

$\int{{{e}^{x}}\left( \frac{x-1}{2{{x}^{2}}} \right)dx}=\frac{1}{2}\int{{{e}^{x}}\left( t+{{t}^{'}} \right)dx}$

Use the integration formula $\int{{{e}^{x}}\left[ f\left( x \right)+f'\left( x \right) \right]}dx={{e}^{x}}f\left( x \right)+C$ in the above equation.

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

$\Rightarrow \int{{{e}^{x}}\left( \frac{x-1}{2{{x}^{2}}} \right)dx}=\frac{{{e}^{x}}}{2x}+C$

Hence the value of $\int{{{e}^{x}}\left( \frac{x-1}{2{{x}^{2}}} \right)dx}$ is $\frac{{{e}^{x}}}{2x}+C$.


(ix). $\int{\sqrt{2ax-{{x}^{2}}}}dx$

Ans: - Simplify the given expression as 

$\int{\sqrt{2ax-{{x}^{2}}}}dx=\int{\sqrt{2\left( a \right)x-{{x}^{2}}+{{a}^{2}}-{{a}^{2}}}}dx$

$\Rightarrow \int{\sqrt{2ax-{{x}^{2}}}}dx=\int{\sqrt{{{a}^{2}}-{{\left( x-a \right)}^{2}}}}dx$

Use the integration formula $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right)+C$ in the above equation.

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

Hence the value of $\int{\sqrt{2ax-{{x}^{2}}}}dx$ is $\frac{x-a}{2}\sqrt{2ax-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x-a}{a} \right)+C$.


(x). $\int{{{e}^{x}}\frac{\left( {{x}^{2}}+1 \right)}{{{\left( x+1 \right)}^{2}}}dx}$

Ans: - Simplify the given expression as 

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

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

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

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

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

If $f\left( x \right)=\frac{1}{x+1}$, then $f'\left( x \right)=\frac{-1}{{{\left( x+1 \right)}^{2}}}$. So, use the formula $\int{{{e}^{x}}\left[ f\left( x \right)+f'\left( x \right) \right]}dx={{e}^{x}}f\left( x \right)+C$ in the above equation.

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

$\Rightarrow \int{{{e}^{x}}\frac{\left( {{x}^{2}}+1 \right)}{{{\left( x+1 \right)}^{2}}}dx}={{e}^{x}}\left[ 1-\frac{2}{x+1} \right]+C$

$\Rightarrow \int{{{e}^{x}}\frac{\left( {{x}^{2}}+1 \right)}{{{\left( x+1 \right)}^{2}}}dx}={{e}^{x}}\left( \frac{x-1}{x+1} \right)+C$

Hence the value of $\int{{{e}^{x}}\frac{\left( {{x}^{2}}+1 \right)}{{{\left( x+1 \right)}^{2}}}dx}$ is ${{e}^{x}}\left( \frac{x-1}{x+1} \right)+C$.


(xi). $\int{{{e}^{x}}\frac{\left( 2+\sin 2x \right)}{\left( 1+\cos 2x \right)}dx}$

Ans: - Use the trigonometric formulas $\sin 2x=2\sin x\cos x$ and $\cos 2x=2{{\cos }^{2}}x-1$ in the given expression.

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

$\Rightarrow \int{{{e}^{x}}\frac{\left( 2+\sin 2x \right)}{\left( 1+\cos 2x \right)}dx}=\int{{{e}^{x}}\left[ {{\sec }^{2}}x+\tan x \right]dx}$

If $f\left( x \right)=\tan x$then $f'\left( x \right)={{\sec }^{2}}x$. So, use the formula $\int{{{e}^{x}}\left[ f\left( x \right)+f'\left( x \right) \right]}dx={{e}^{x}}f\left( x \right)+C$ in the above equation.

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

Hence the value of $\int{{{e}^{x}}\frac{\left( 2+\sin 2x \right)}{\left( 1+\cos 2x \right)}dx}$ is ${{e}^{x}}\tan x+C$.


(xii). $\int{\left\{ \log \left( \log x \right)+\frac{1}{{{\left( \log x \right)}^{2}}} \right\}dx}$

Ans: - Consider $I=\int{\log \left( \log x \right)dx}$

Apply integration by parts.

$I=\log \left( \log x \right)\int{1dx}-\int{\left[ \left( \log \left( \log x \right)'\times \int{1dx} \right) \right]dx}$

$\Rightarrow I=x\log \left( \log x \right)-\int{\frac{1}{\log x}\times \frac{1}{x}\times xdx}$

$\Rightarrow I=x\log \left( \log x \right)-\int{{{\left( \log x \right)}^{-1}}dx}$

Consider ${{I}_{1}}=\int{{{\left( \log x \right)}^{-1}}dx}$

Apply integration by parts.

$I=x\log \left( \log x \right)-\left[ {{\left( \log x \right)}^{-1}}\int{1dx}-\int{\left[ \frac{d}{dx}\left( {{\left( \log x \right)}^{-1}} \right)\int{1dx} \right]dx} \right]$

$I=x\log \left( \log x \right)-\left[ \frac{x}{\log x}-\int{-{{\left( \log x \right)}^{-2}}\times \frac{1}{x}\times xdx} \right]$

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

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

Hence the value of $\int{\left\{ \log \left( \log x \right)+\frac{1}{{{\left( \log x \right)}^{2}}} \right\}dx}$ is $x\log \left( \log x \right)-\frac{x}{\log x}+C$.


(xiii). $\int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}$

Ans: - Let $6x+5=A.\frac{d}{dx}\left( 6-{{x}^{2}}+x \right)+B$

$\Rightarrow 6x+5=A\left( -2x+1 \right)+B$

$\Rightarrow 6x+5=-2Ax+A+B$

Equating and solving for coefficients, then we will get $A=-3$ and $B=8$.

Now the given expression can be written as 

$\int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=\int{\left[ -3\left( -2x+1 \right)+8 \right]\sqrt{6-{{x}^{2}}+x}dx}$

$\Rightarrow \int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=-3\int{\left( -2x+1 \right)\sqrt{6+x-{{x}^{2}}}}+8\int{\sqrt{6+x-{{x}^{2}}}}dx$

Assume $6+x-{{x}^{2}}=t\Leftrightarrow \left( 1-2x \right)dx=dt$ in the above equation.

$\int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=-3\int{\sqrt{t}dt}+8\int{\sqrt{6+2\left( \frac{1}{2} \right)x-{{x}^{2}}-{{\left( \frac{1}{2} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}}}dx$

$\Rightarrow \int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=-3\times \frac{{{t}^{\frac{3}{2}}}}{\frac{3}{2}}+8\int{\sqrt{{{\left( \frac{5}{2} \right)}^{2}}-{{\left( x-\frac{1}{2} \right)}^{2}}}}dx$

Use the integration formula $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{2} \right)+C$ and substitute $t=6+x-{{x}^{2}}$ in the above equation.

$\int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=-2{{\left( 6+x-{{x}^{2}} \right)}^{\frac{3}{2}}}+8\left[ \frac{x-\frac{1}{2}}{2}\sqrt{6+x-{{x}^{2}}}+\frac{{{\left( \frac{5}{2} \right)}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x-\frac{1}{2}}{2} \right) \right]+C$

$\Rightarrow \int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}=-2{{\left( 6+x-{{x}^{2}} \right)}^{\frac{3}{2}}}+8\left[ \frac{2x-1}{4}\sqrt{6+x-{{x}^{2}}}+\frac{25}{8}{{\sin }^{-1}}\left( \frac{2x-1}{4} \right) \right]+C$

Hence the value of $\int{\left( 6x+5 \right)\sqrt{6+x-{{x}^{2}}}dx}$ is $-2{{\left( 6+x-{{x}^{2}} \right)}^{\frac{3}{2}}}+8\left[ \frac{2x-1}{4}\sqrt{6+x-{{x}^{2}}}+\frac{25}{8}{{\sin }^{-1}}\left( \frac{2x-1}{4} \right) \right]+C$.


(xiv). $\int{\left( x-2 \right)\sqrt{\frac{x+3}{x-3}}dx}$

Ans:

$\int{\left( \text{x-2} \right)\sqrt{\frac{\text{x+3}}{\text{x-3}}}\text{dx}}=\frac{\sqrt{{{x}^{2}}-9}\left[ \left( x+2 \right)\sqrt{9-{{x}^{2}}}-6{{\sin }^{-1}}\left( \frac{\sqrt{3-x}}{\sqrt{6}} \right) \right]}{2\sqrt{9-{{x}^{2}}}}+C$, where $C$ is an integration constant.


(xv). $\int{\left( 2x-5 \right)\sqrt{{{x}^{2}}-4x+3}}dx$

Ans: Simplify the given expression as  

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

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

Let $u={{x}^{2}}-4x+3\Leftrightarrow du=\left( 2x-4 \right)dx$, now the above equation is modified as 

$\int{\left( 2x-5 \right)\sqrt{{{x}^{2}}-4x+3}}dx=\int{\sqrt{u}du}-\int{\sqrt{{{x}^{2}}-2\left( 2 \right)x+{{2}^{2}}-{{2}^{2}}+3}}dx$

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

Use the integration formula $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C$ and substitute $u={{x}^{2}}-4x+3$ in the above equation.

$\int{\left( 2x-5 \right)\sqrt{{{x}^{2}}-4x+3}}dx=\frac{2}{3}{{\left( {{x}^{2}}-4x+3 \right)}^{\frac{3}{2}}}-\left[ \frac{x-2}{2}\sqrt{{{x}^{2}}-4x+3}+\frac{1}{2}\log \left| x-2+\sqrt{{{x}^{2}}-4x+3} \right| \right]+C$

$\Rightarrow \int{\left( 2x-5 \right)\sqrt{{{x}^{2}}-4x+3}}dx=\frac{2}{3}{{\left( {{x}^{2}}-4x+3 \right)}^{\frac{3}{2}}}-\left( \frac{x-2}{2} \right)\sqrt{{{x}^{2}}-4x+3}+\frac{1}{2}\log \left| x-2+\sqrt{{{x}^{2}}-4x+3} \right|+C$

Hence the value of $\int{\left( 2x-5 \right)\sqrt{{{x}^{2}}-4x+3}}dx$ is $\frac{2}{3}{{\left( {{x}^{2}}-4x+3 \right)}^{\frac{3}{2}}}-\left( \frac{x-2}{2} \right)\sqrt{{{x}^{2}}-4x+3}+\frac{1}{2}\log \left| x-2+\sqrt{{{x}^{2}}-4x+3} \right|+C$.


(xiv). $\int{\sqrt{{{x}^{2}}-4x+8}}dx$

Ans: - Simplify the above equation as 

$\int{\sqrt{{{x}^{2}}-4x+8}}dx=\int{\sqrt{{{x}^{2}}-2\left( 2 \right)x+{{2}^{2}}-{{2}^{2}}+8}}dx$

$\Rightarrow \int{\sqrt{{{x}^{2}}-4x+8}}dx=\int{\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}}dx$

Use the integration formula $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C$ in the above equation.

$\int{\sqrt{{{x}^{2}}-4x+8}}dx=\frac{x-2}{2}\sqrt{{{x}^{2}}-4x+8}+\frac{{{2}^{2}}}{2}\log \left| x-2+\sqrt{{{x}^{2}}-4x+8} \right|+C$

Hence the value of $\int{\sqrt{{{x}^{2}}-4x+8}}dx$ is $\left( \frac{x-2}{2} \right)\sqrt{{{x}^{2}}-4x+8}+2\log \left| \left( x-2 \right)+\sqrt{{{x}^{2}}-4x+8} \right|+C$.


  1. Evaluate the following definite integrals:

(i). \[\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x+\cos x}{9+16\sin 2x}dx}\]

Ans: Given integral is – \[\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x+\cos x}{9+16\sin 2x}dx}\]

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

$\Rightarrow (\cos x+\sin x)dx=dt$

For the limits –

  • When $x=0\Rightarrow t=-1$

  • When $x=\frac{\pi }{4}\Rightarrow t=0$

Now, 

\[{{\left( \sin x-\cos x \right)}^{2}}={{t}^{2}}\]

\[\Rightarrow \left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)={{t}^{2}}\]

It is known that, \[\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)=1\] and $\sin 2x=2\sin x\cos x$

\[\Rightarrow \left( 1-\sin 2x \right)={{t}^{2}}\]

\[\Rightarrow 1-{{t}^{2}}=\sin 2x\]

Using the above equations, the integral changes to – 

\[\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x+\cos x}{9+16\sin 2x}dx}=\int\limits_{-1}^{0}{\frac{dt}{9+16(1-{{t}^{2}})}}\]

\[=\int\limits_{-1}^{0}{\frac{dt}{9+16-16{{t}^{2}}}}\]

\[=\int\limits_{-1}^{0}{\frac{dt}{25-16{{t}^{2}}}}\]

\[=\int\limits_{-1}^{0}{\frac{dt}{{{(5)}^{2}}-{{(4t)}^{2}}}}\]

Using – $\int{\frac{dx}{{{a}^{2}}-{{x}^{2}}}=\frac{1}{2a}\log \left| \frac{a+x}{a-x} \right|+C}$

\[=\frac{1}{4}\left[ \frac{1}{2\times 5}\log \left| \frac{5+4t}{5-4t} \right| \right]_{-1}^{0}\]

\[=\frac{1}{4}\left[ \frac{1}{10}\log \left| \frac{5}{5} \right|-\frac{1}{10}\log \left| \frac{5-4}{5+4} \right| \right]\]

\[=\frac{1}{4}\times \frac{1}{10}\left[ \log \left| 1 \right|-\log \left| \frac{1}{9} \right| \right]\]

\[=\frac{1}{40}\left[ \log \left| 1 \right|-\log \left| 1 \right|+\log \left| 9 \right| \right]\]

\[=\frac{1}{40}\log \left| 9 \right|\]

Thus, \[\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x+\cos x}{9+16\sin 2x}dx}=\frac{1}{40}\log \left| 9 \right|\]


(ii). \[\int\limits_{0}^{\frac{\pi }{2}}{\cos 2x\log \sin xdx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{2}}{\cos 2x\log \sin xdx}\]

Using integration by parts – \[\int{uvdx}=u\int{vdx}-\int{u'\left( \int{vdx} \right)dx}\]

Here, 

$u=\log \sin x\Rightarrow u'=\frac{1}{\sin x}\times \cos x$

$v=\cos 2x\Rightarrow \int{vdx}=\frac{1}{2}\sin 2x$

The integral becomes – 

\[\int\limits_{0}^{\frac{\pi }{2}}{\cos 2x\log \sin xdx}=\left[ \log \left( \sin x \right)\times \frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\left( \frac{1}{\sin x}\cos x\times \sin 2x \right)dx}\]

It is known that – $\sin 2x=2\sin x\cos x$

\[=\left[ \log \left( \sin x \right)\times \frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\left( \frac{1}{\sin x}\cos x\times 2\sin x\cos x \right)dx}\]

\[=\left[ \log \left( \sin x \right)\times \frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}-\int\limits_{0}^{\frac{\pi }{2}}{\left( {{\cos }^{2}}x \right)dx}\]

It is known that – $\cos 2x=2{{\cos }^{2}}x-1$

\[=\left[ \log \left( \sin x \right)\times \frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\left( 1+\cos 2x \right)dx}\]

\[=\left[ \log \left( \sin x \right)\times \frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}-\frac{1}{2}\left[ x+\frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}\]

Substituting the limits – 

\[=\left[ \log \left( \sin \frac{\pi }{2} \right)\times \frac{1}{2}\sin \pi -\log \left( \sin 0 \right)\times \frac{1}{2}\sin 0 \right]-\frac{1}{2}\left[ \frac{\pi }{2}+\frac{1}{2}\sin \pi -0-\frac{1}{2}\sin 0 \right]\]

\[=\left[ 0-0 \right]-\frac{1}{2}\left[ \frac{\pi }{2}+0-0-0 \right]\]

$\Rightarrow I=-\frac{\pi }{4}$

Thus, \[\int\limits_{0}^{\frac{\pi }{2}}{\cos 2x\log \sin xdx}=-\frac{\pi }{4}\]


(iii). \[\int\limits_{0}^{1}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}dx}\]

Ans: Given integral – \[\int\limits_{0}^{1}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}dx}\]

Say, ${{x}^{2}}=t\Rightarrow 2xdx=dt$

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=1\Rightarrow t=1$

The integral becomes – 

\[\int\limits_{0}^{1}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}dx}=\frac{1}{2}\int\limits_{0}^{1}{\sqrt{\frac{1-t}{1+t}}dt}\]

Multiply the numerator and denominator by $\sqrt{1-t}$

\[=\frac{1}{2}\int\limits_{0}^{1}{\sqrt{\frac{\left( 1-t \right)\left( 1-t \right)}{\left( 1+t \right)\left( 1-t \right)}}dt}\]

$=\frac{1}{2}\int\limits_{0}^{1}{\frac{1-t}{\sqrt{1-{{t}^{2}}}}dt}$

$=\frac{1}{2}\left[ \int\limits_{0}^{1}{\frac{1}{\sqrt{1-{{t}^{2}}}}dt-\int\limits_{0}^{1}{\frac{t}{\sqrt{1-{{t}^{2}}}}dt}} \right]$

Integrating the first term – 

$\int\limits_{0}^{1}{\frac{1}{\sqrt{1-{{t}^{2}}}}dt=\left[ {{\sin }^{-1}}t \right]_{0}^{1}}$

$=\left[ {{\sin }^{-1}}1-{{\sin }^{-1}}0 \right]$

$=\frac{\pi }{2}-0$

$\Rightarrow \int\limits_{0}^{1}{\frac{1}{\sqrt{1-{{t}^{2}}}}dt=\frac{\pi }{2}}$

Integrating the second term – $\int\limits_{0}^{1}{\frac{t}{\sqrt{1-{{t}^{2}}}}dt}$

Say, $1-{{t}^{2}}=u\Rightarrow -2tdt=du$

The limits – 

  • When $t=0\Rightarrow u=1$

  • When $t=1\Rightarrow u=0$

$\therefore \int\limits_{0}^{1}{\frac{t}{\sqrt{1-{{t}^{2}}}}dt}=-\frac{1}{2}\int\limits_{1}^{0}{\frac{du}{\sqrt{u}}}$

Interchanging the limits – 

$=\frac{1}{2}\int\limits_{0}^{1}{\frac{du}{\sqrt{u}}}$

$=\frac{1}{2}\left[ \frac{1}{\left( \frac{1}{2} \right)}\sqrt{u} \right]_{0}^{1}$

$=\sqrt{1}-0$

$\int\limits_{0}^{1}{\frac{t}{\sqrt{1-{{t}^{2}}}}dt}=1$

Using them – 

$\int\limits_{0}^{1}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}dx}=\frac{1}{2}\left[ \int\limits_{0}^{1}{\frac{1}{\sqrt{1-{{t}^{2}}}}dt-\int\limits_{0}^{1}{\frac{t}{\sqrt{1-{{t}^{2}}}}dt}} \right]$

$=\frac{1}{2}\left[ \frac{\pi }{2}-1 \right]$

$=\frac{\pi }{4}-\frac{1}{2}$

Thus, \[\int\limits_{0}^{1}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}dx}=\frac{\pi }{4}-\frac{1}{2}\]


(iii). \[\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}\]

\[\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}=\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}dx}\]

Say, ${{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=\frac{1}{\sqrt{2}}\Rightarrow t=\frac{\pi }{4}$

Also, $x=\sin t\Rightarrow {{x}^{2}}={{\sin }^{2}}t$

So, \[\left( 1-{{x}^{2}} \right)=\left( 1-{{\sin }^{2}}t \right)\Rightarrow \left( 1-{{x}^{2}} \right)={{\cos }^{2}}t\]

The integral becomes – 

\[\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}=\int\limits_{0}^{\frac{\pi }{4}}{\frac{t}{{{\cos }^{2}}t}dt}\]

\[=\int\limits_{0}^{\frac{\pi }{4}}{t{{\sec }^{2}}tdt}\]

Using integration by parts – \[\int{uvdx}=u\int{vdx}-\int{u'\left( \int{vdx} \right)dx}\]

Here, 

$u=t\Rightarrow u'=1$

$v={{\sec }^{2}}t\Rightarrow \int{vdx}=\tan t$

The integral becomes – 

\[\int\limits_{0}^{\frac{\pi }{4}}{t{{\sec }^{2}}tdt}=\left[ t\tan t \right]_{0}^{\frac{\pi }{4}}-\int\limits_{0}^{\frac{\pi }{4}}{\left( 1\times \tan t \right)dt}\]

\[=\left[ t\tan t \right]_{0}^{\frac{\pi }{4}}-\int\limits_{0}^{\frac{\pi }{4}}{\left( \tan t \right)dt}\]

\[=\left[ t\tan t \right]_{0}^{\frac{\pi }{4}}-\left[ -\log \left( \cos t \right) \right]_{0}^{\frac{\pi }{4}}\]

Applying the limits,

\[=\left[ \frac{\pi }{4}\tan \frac{\pi }{4}-0 \right]+\left[ \log \left( \cos \frac{\pi }{4} \right)-\log \left( \cos 0 \right) \right]\]

\[=\left[ \frac{\pi }{4} \right]+\left[ \log \left( \frac{1}{\sqrt{2}} \right)-\log \left( 1 \right) \right]\]

\[=\frac{\pi }{4}+\left[ \log 1-\log \sqrt{2}-\log 1 \right]\]

\[=\frac{\pi }{4}+\left[ -\log \sqrt{2} \right]\]

\[=\frac{\pi }{4}-\frac{1}{2}\log 2\]

Thus, \[\int\limits_{0}^{\frac{1}{\sqrt{2}}}{\frac{{{\sin }^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}=\frac{\pi }{4}-\frac{1}{2}\log 2\]


(iv). \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\]

\[\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{2\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\]

Divide the numerator and denominator by ${{\cos }^{4}}x$

\[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\frac{2\sin x\cos x}{{{\cos }^{4}}x}}{\frac{{{\sin }^{4}}x}{{{\cos }^{4}}x}+\frac{{{\cos }^{4}}x}{{{\cos }^{4}}x}}dx}\]

Simplify,

\[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{2\tan x{{\sec }^{2}}x}{{{\tan }^{4}}x+1}dx}\]

Say, $\tan x=t\Rightarrow {{\sec }^{2}}xdx=dt$

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=\frac{\pi }{2}\Rightarrow t=\infty $

The integral becomes – 

\[=\int\limits_{0}^{\infty }{\frac{2tdt}{{{t}^{4}}+1}}\]

Say, ${{\operatorname{t}}^{2}}=u\Rightarrow 2tdt=du$

The limits – 

  • When $t=0\Rightarrow u=0$

  • When $t=\infty \Rightarrow u=\infty $

The integral becomes – 

\[=\int\limits_{0}^{\infty }{\frac{du}{{{u}^{2}}+1}}\]

$=\left[ {{\tan }^{-1}}u \right]_{0}^{\infty }$

$=\left[ {{\tan }^{-1}}\infty -{{\tan }^{-1}}0 \right]$

$=\frac{\pi }{2}-0$

\[\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\frac{\pi }{2}\]

Thus, \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\frac{\pi }{2}\]


(v). \[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}\]

Ans: Given integral – \[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}\]

\[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\int\limits_{1}^{2}{\left\{ 5-\frac{20x+15}{{{x}^{2}}+4x+3} \right\}dx}\]

\[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\left[ 5x \right]_{1}^{2}-\int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}\to (1)\]                         

Solving \[\int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}\] ,

Let \[20x+15=A\frac{d}{dx}\left( {{x}^{2}}+4x+3 \right)+B\]

Equating the coefficients of \[x\] and constant term we get,

\[A=10,B=-25\]

Let \[{{x}^{2}}+4x+3=t\]

Differentiating this we get,

\[\left( 2x+4 \right)dx=dt\]

The integral becomes –

\[10\int{\frac{dt}{t}-25\int{\frac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}\]

\[10\int{\frac{dt}{t}-25\int{\frac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}=10\log t-25\left[ \frac{1}{2}\log \left( \frac{x+2-1}{x+2+1} \right) \right]\]

\[\Rightarrow \int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}=\left[ 10\log \left( {{x}^{2}}+4x+3 \right)-25\left[ \frac{1}{2}\log \left( \frac{x+1}{x+3} \right) \right] \right]_{1}^{2}\]

\[\Rightarrow \int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 15-10\log 8-25\left[ \frac{1}{2}\log \frac{3}{5}-\frac{1}{2}\log \frac{2}{4} \right]\]

\[\Rightarrow \int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 5+10\log 3-10\log 4-10\log 2-\frac{25}{2}\left[ \log 3-\log 5-\log 2+\log 4 \right]\]\[\Rightarrow \int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}=\frac{45}{2}\log 5-\frac{45}{2}\log 4-\frac{5}{2}\log 3+\frac{5}{2}\log 2\]

\[\therefore \int\limits_{1}^{2}{\frac{20x+15}{{{x}^{2}}+4x+3}dx}=\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}\]

Substituting in $(1)$ we get,

\[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\left[ \frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2} \right]\]

Thus,  \[\int\limits_{1}^{2}{\frac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\frac{5}{2}\left[ 9\log \frac{5}{4}-\log \frac{3}{2} \right]\]


(vi). \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+\sin x}{1+\cos x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+\sin x}{1+\cos x}dx}\]

It is known that –

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$1+\cos x=2{{\cos }^{2}}\frac{x}{2}$

So,

\[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+\sin x}{1+\cos x}dx}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}dx}\]

Simplify,

\[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{x}{2{{\cos }^{2}}\frac{x}{2}}dx}+\int\limits_{0}^{\frac{\pi }{2}}{\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}dx}\]

\[=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{x{{\sec }^{2}}\frac{x}{2}dx}+\int\limits_{0}^{\frac{\pi }{2}}{\tan \frac{x}{2}dx}\]

Evaluating the first term – \[\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{x{{\sec }^{2}}\frac{x}{2}dx}\]

Using integration by parts – \[\int{uvdx}=u\int{vdx}-\int{u'\left( \int{vdx} \right)dx}\]

Here, 

$u=x\Rightarrow u'=1$

$v={{\sec }^{2}}\frac{x}{2}\Rightarrow \int{vdx}=2\tan \frac{x}{2}$

The integral becomes – 

\[\int\limits_{0}^{\frac{\pi }{2}}{x{{\sec }^{2}}\frac{x}{2}dt}=\left[ 2x\tan \frac{x}{2} \right]_{0}^{\frac{\pi }{2}}-\int\limits_{0}^{\frac{\pi }{2}}{\left( 1\times 2\tan \frac{x}{2} \right)dt}\]

\[=\left[ 2x\tan \frac{x}{2} \right]_{0}^{\frac{\pi }{2}}-2\int\limits_{0}^{\frac{\pi }{2}}{\left( \tan \frac{x}{2} \right)dt}\]

\[=\left[ 2x\tan \frac{x}{2} \right]_{0}^{\frac{\pi }{2}}-2\int\limits_{0}^{\frac{\pi }{2}}{\left( \tan \frac{x}{2} \right)dt}\]

Applying the limits,

\[=\left[ \pi \tan \frac{\pi }{4}-0 \right]-2\int\limits_{0}^{\frac{\pi }{2}}{\left( \tan \frac{x}{2} \right)dt}\]

\[=\left[ \pi  \right]-2\int\limits_{0}^{\frac{\pi }{2}}{\left( \tan \frac{x}{2} \right)dt}\]

Using this,

\[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+\sin x}{1+\cos x}dx}=\left[ \frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{x{{\sec }^{2}}\frac{x}{2}dx} \right]+\int\limits_{0}^{\frac{\pi }{2}}{\tan \frac{x}{2}dx}\]

\[=\frac{1}{2}\left[ \left[ \pi  \right]-2\int\limits_{0}^{\frac{\pi }{2}}{\left( \tan \frac{x}{2} \right)dt} \right]+\int\limits_{0}^{\frac{\pi }{2}}{\tan \frac{x}{2}dx}\]

\[=\frac{\pi }{2}-\int\limits_{0}^{\frac{\pi }{2}}{\tan \frac{x}{2}dx}+\int\limits_{0}^{\frac{\pi }{2}}{\tan \frac{x}{2}dx}\]

$=\frac{\pi }{2}$

Thus, \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x+\sin x}{1+\cos x}dx}=\frac{\pi }{2}\]


  1.  Evaluate:

(i). \[\int\limits_{1}^{3}{\left\{ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right\}dx}\]

Ans: Given integral – \[\int\limits_{1}^{3}{\left\{ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right\}dx}\]

Say, \[I=\int\limits_{1}^{3}{\left\{ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right\}dx}\]

Splitting the integral – 

\[\Rightarrow I=\int\limits_{1}^{3}{\left| x-1 \right|dx}+\int\limits_{1}^{3}{\left| x-2 \right|dx}+\int\limits_{1}^{3}{\left| x-3 \right|dx}\]

i.e. $I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}$

with – 

\[{{I}_{1}}=\int\limits_{1}^{3}{\left| x-1 \right|dx}\]

\[{{I}_{2}}=\int\limits_{1}^{3}{\left| x-2 \right|dx}\]

\[{{I}_{3}}=\int\limits_{1}^{3}{\left| x-3 \right|dx}\]

Now, 

\[{{I}_{1}}=\int\limits_{1}^{3}{\left| x-1 \right|dx}\]

Here, $\left( x-1 \right)\ge 0\text{ for 1}\le \text{x}\le \text{3}$

\[\Rightarrow {{I}_{1}}=\int\limits_{1}^{3}{(x-1)dx}\]

\[=\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{3}\]

\[=\left[ \frac{{{3}^{2}}}{2}-3-\frac{{{1}^{2}}}{2}+1 \right]\]

\[=\left[ \frac{9}{2}-\frac{1}{2}-2 \right]\]

\[=\left[ \frac{8}{2}-2 \right]\]

$=4-2$

$=2$

\[\Rightarrow {{I}_{1}}=2\]

\[{{I}_{2}}=\int\limits_{1}^{3}{\left| x-2 \right|dx}\]

Here, $\left( x-2 \right)\ge 0\text{ for 2}\le \text{x}\le \text{3 and }\left( x-2 \right)\le 0\text{ for 1}\le \text{x}\le 2\text{ }$

\[\Rightarrow {{I}_{2}}=\int\limits_{1}^{2}{-(x-2)dx}+\int\limits_{2}^{3}{(x-2)dx}\]

\[=-\left[ \frac{{{x}^{2}}}{2}-2x \right]_{1}^{2}+\left[ \frac{{{x}^{2}}}{2}-2x \right]_{2}^{3}\]

\[=-\left[ \frac{{{2}^{2}}}{2}-4-\frac{{{1}^{2}}}{2}+2 \right]+\left[ \frac{{{3}^{2}}}{2}-6-\frac{{{2}^{2}}}{2}+4 \right]\]

\[=-\left[ \frac{4}{2}-\frac{1}{2}-2 \right]+\left[ \frac{9}{2}-\frac{4}{2}-2 \right]\]

\[=-\left[ \frac{3}{2}-2 \right]+\left[ \frac{5}{2}-2 \right]\]

\[=-\frac{3}{2}+2+\frac{5}{2}-2\]

$=\frac{2}{2}$

$=1$

\[\Rightarrow {{I}_{2}}=1\]

\[{{I}_{3}}=\int\limits_{1}^{3}{\left| x-3 \right|dx}\]

Here, $\left( x-3 \right)\le 0\text{ for 1}\le \text{x}\le \text{3}$

\[\Rightarrow {{I}_{3}}=\int\limits_{1}^{3}{-(x-3)dx}\]

\[=-\left[ \frac{{{x}^{2}}}{2}-3x \right]_{1}^{3}\]

\[=-\left[ \frac{{{3}^{2}}}{2}-9-\frac{{{1}^{2}}}{2}+3 \right]\]

\[=-\left[ \frac{9}{2}-\frac{1}{2}-6 \right]\]

\[=-\left[ \frac{8}{2}-6 \right]\]

$=-4+6$

$=2$

\[\Rightarrow {{I}_{3}}=2\]

$I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}$

$\Rightarrow I=2+1+2$

$\Rightarrow I=5$

Thus, \[\int\limits_{1}^{3}{\left\{ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right\}dx}=5\]


(ii). \[\int\limits_{0}^{\pi }{\frac{x}{1+\sin x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\pi }{\frac{x}{1+\sin x}dx}\]

Say, \[I=\int\limits_{0}^{\pi }{\frac{x}{1+\sin x}dx}\to (1)\]

Using – $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

\[\Rightarrow I=\int\limits_{0}^{\pi }{\frac{(\pi -x)}{1+\sin (\pi -x)}dx}\]

It is known that – $\sin (\pi -x)=\sin x$

\[\Rightarrow I=\int\limits_{0}^{\pi }{\frac{(\pi -x)}{1+\sin x}dx}\to (2)\]

Add equations – $(1)+(2)$

\[I+I=\int\limits_{0}^{\pi }{\frac{x}{1+\sin x}dx}+\int\limits_{0}^{\pi }{\frac{(\pi -x)}{1+\sin x}dx}\]

\[\Rightarrow 2I=\int\limits_{0}^{\pi }{\frac{x+\pi -x}{1+\sin x}dx}\]

\[=\int\limits_{0}^{\pi }{\frac{\pi }{1+\sin x}dx}\]

\[=\pi \int\limits_{0}^{\pi }{\frac{1}{1+\sin x}dx}\]

Multiply and divide by $(1-\sin x)$

\[=\pi \int\limits_{0}^{\pi }{\frac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}dx}\]

\[=\pi \int\limits_{0}^{\pi }{\frac{\left( 1-\sin x \right)}{\left( 1-{{\sin }^{2}}x \right)}dx}\]

It is known that – ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

\[=\pi \int\limits_{0}^{\pi }{\frac{\left( 1-\sin x \right)}{\left( {{\cos }^{2}}x \right)}dx}\]

Simplifying

\[=\pi \int\limits_{0}^{\pi }{\left\{ \left( \frac{1}{{{\cos }^{2}}x} \right)-\left( \frac{\sin x}{{{\cos }^{2}}x} \right) \right\}dx}\]

\[=\pi \int\limits_{0}^{\pi }{\left\{ {{\sec }^{2}}x-\sec x\tan x \right\}dx}\]

\[=\pi \left[ \int\limits_{0}^{\pi }{{{\sec }^{2}}xdx}-\int\limits_{0}^{\pi }{\sec x\tan xdx} \right]\]

$=\pi \left[ \tan x \right]_{0}^{\pi }-\pi \left[ \sec x \right]_{0}^{\pi }$

$=\pi \left[ \tan \pi -\tan 0 \right]-\pi \left[ \sec \pi -\sec 0 \right]$

$=\pi \left[ 0-0 \right]-\pi \left[ -1-1 \right]$

$\Rightarrow 2I=2\pi $

$\Rightarrow I=\pi $

Thus, \[\int\limits_{0}^{\pi }{\frac{x}{1+\sin x}dx}=\pi \]


(iii). \[\int\limits_{0}^{\frac{\pi }{4}}{\log (1+\tan x)dx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{4}}{\log (1+\tan x)dx}\]

Say,$I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left( 1+\tan x \right)dx}\to (1)$

Using – $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

 $\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left[ 1+\tan \left( \frac{\pi }{4}-x \right) \right]dx}$

\[=\int\limits_{0}^{\frac{\pi }{4}}{\log \left[ 1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x} \right]dx}\]

$=\int\limits_{0}^{\frac{\pi }{4}}{\log \left\{ 1+\frac{1-\tan x}{1+\tan x} \right\}dx}$

$=\int\limits_{0}^{\frac{\pi }{4}}{\log \frac{2}{1+\tan x}}dx$

$=\int\limits_{0}^{\frac{\pi }{4}}{\log 2dx}-\int\limits_{0}^{\frac{\pi }{4}}{\log \left( 1+\tan x \right)}dx$

Comparing with $(1)$

$=\int\limits_{0}^{\frac{\pi }{4}}{\log 2dx}-I$

$\Rightarrow 2I=\left[ x\log 2 \right]_{0}^{\frac{\pi }{4}}$

$\Rightarrow 2I=\frac{\pi }{4}\log 2$

$\Rightarrow I=\frac{\pi }{8}\log 2$

Thus, \[\int\limits_{0}^{\frac{\pi }{4}}{\log (1+\tan x)dx}=\frac{\pi }{8}\log 2\]


(iv). \[\int\limits_{0}^{\frac{\pi }{2}}{\log \sin xdx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{2}}{\log \sin xdx}\]

Say, \[I=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin xdx}\to (1)\]

Using – $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

\[\Rightarrow I=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin \left( \frac{\pi }{2}-x \right)dx}\]

\[\Rightarrow I=\int\limits_{0}^{\frac{\pi }{2}}{\log \cos xdx}\to (2)\]

Adding equations – $(1)+(2)$

\[I+I=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin xdx}+\int\limits_{0}^{\frac{\pi }{2}}{\log \cos xdx}\]

\[\Rightarrow 2I=\int\limits_{0}^{\frac{\pi }{2}}{\left( \log \sin x+\log \cos x \right)dx}\]

Add and subtract $\log 2$

\[=\int\limits_{0}^{\frac{\pi }{2}}{\left( \log \sin x+\log \cos x+\log 2-\log 2 \right)dx}\]

It is known that – $\log m+\log n=\log mn$

\[=\int\limits_{0}^{\frac{\pi }{2}}{\left( \log 2\sin x\cos x-\log 2 \right)dx}\]

\[=\int\limits_{0}^{\frac{\pi }{2}}{\left( \log \sin 2x-\log 2 \right)dx}\]

\[\therefore 2I=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin 2xdx}-\int\limits_{0}^{\frac{\pi }{2}}{\log 2dx}\]

Say, $2I={{I}_{1}}-{{I}_{2}}$

With – 

\[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin 2xdx}\]

\[{{I}_{2}}=\int\limits_{0}^{\frac{\pi }{2}}{\log 2dx}\]

Now, 

\[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin 2xdx}\]

Put $2x=t\Rightarrow 2dx=dt$

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=\frac{\pi }{2}\Rightarrow t=\pi $

\[\Rightarrow {{I}_{1}}=\frac{1}{2}\int\limits_{0}^{\pi }{\log \sin tdt}\]

Using – $\int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx},\text{ when f(2a}-x)=f(x)$

Here, $\sin (\pi -t)=\sin t$

\[\Rightarrow {{I}_{1}}=\frac{1}{2}\times 2\int\limits_{0}^{\frac{\pi }{2}}{\log \sin tdt}\]

\[\Rightarrow {{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\log \sin tdt}\]

Comparing with equation $(1)$

\[\Rightarrow {{I}_{1}}=I\]

Now, \[{{I}_{2}}=\int\limits_{0}^{\frac{\pi }{2}}{\log 2dx}\]

\[=\log 2\int\limits_{0}^{\frac{\pi }{2}}{dx}\]

$=\log 2\left[ x \right]_{0}^{\frac{\pi }{2}}$

$=\log 2\left[ \frac{\pi }{2}-0 \right]$

$\Rightarrow {{I}_{2}}=\frac{\pi }{2}\log 2$

Substituting them – 

$2I={{I}_{1}}-{{I}_{2}}$

$2I=I-\frac{\pi }{2}\log 2$

$\Rightarrow I=-\frac{\pi }{2}\log 2$

Thus, \[\int\limits_{0}^{\frac{\pi }{2}}{\log \sin xdx}=-\frac{\pi }{2}\log 2\]


(v). \[\int\limits_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\]

Say, \[I=\int\limits_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\to (1)\]

Using – $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

\[\therefore I=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)\sin \left( \pi -x \right)}{1+{{\cos }^{2}}\left( \pi -x \right)}dx}\]

Here, $\sin (\pi -x)=\sin x$ and $\cos (\pi -x)=-\cos x\Rightarrow {{\cos }^{2}}(\pi -x)={{\cos }^{2}}x$

\[=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)\sin x}{1+{{\cos }^{2}}x}dx}\]

\[=\int\limits_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx}-\int\limits_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}\]

Comparing with equation $(1)$

\[\Rightarrow I=\int\limits_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx}-I\]

\[\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx}\]

\[\Rightarrow I=\frac{\pi }{2}\int\limits_{0}^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}dx}\]

Say, $\cos x=t\Rightarrow -\sin xdx=dt$

The limits – 

  • When $x=0\Rightarrow t=1$

  • When $x=\pi \Rightarrow t=-1$

The integral becomes –

\[\Rightarrow I=-\frac{\pi }{2}\int\limits_{1}^{-1}{\frac{dt}{1+{{t}^{2}}}}\]

Interchanging the limits,

\[\Rightarrow I=\frac{\pi }{2}\int\limits_{-1}^{1}{\frac{dt}{1+{{t}^{2}}}}\]

Using – $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx},\text{ when f(}x)\text{ is even}$

Here, \[\frac{1}{1+{{t}^{2}}}\] is an even function.

\[\Rightarrow I=\frac{\pi }{2}\times 2\int\limits_{0}^{1}{\frac{dt}{1+{{t}^{2}}}}\]

\[=\pi \int\limits_{0}^{1}{\frac{dt}{1+{{t}^{2}}}}\]

$=\pi \left[ {{\tan }^{-1}}t \right]_{0}^{1}$

$=\pi \left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]$

$=\pi \left[ \frac{\pi }{4}-0 \right]$

$\Rightarrow I=\frac{{{\pi }^{2}}}{4}$

Thus, \[\int\limits_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}=\frac{{{\pi }^{2}}}{4}\]


(vi). \[\int\limits_{-2}^{2}{f(x)dx\text{ where }f(x)=\left\{ \begin{align} & 2x-{{x}^{3}}\text{ when }-2\le x<-1 \\ & {{x}^{3}}-3x+2\text{ when }-1\le x<1 \\ & 3x-2\text{ when }1\le x<2 \\ \end{align} \right.}\]

Ans: Given integral – \[\int\limits_{-2}^{2}{f(x)dx\text{ where }f(x)=\left\{ \begin{align} & 2x-{{x}^{3}}\text{ when }-2\le x<-1 \\ & {{x}^{3}}-3x+2\text{ when }-1\le x<1 \\ & 3x-2\text{ when }1\le x<2 \\ \end{align} \right.}\]

Splitting based on its intervals –

\[\int\limits_{-2}^{2}{f(x)dx=\int\limits_{-2}^{-1}{\left( 2x-{{x}^{3}} \right)}dx+\int\limits_{-1}^{1}{\left( {{x}^{3}}-3x+2 \right)}dx+\int\limits_{1}^{2}{\left( 3x-2 \right)}dx}\]

\[=\left[ 2\frac{{{x}^{2}}}{2}-\frac{{{x}^{4}}}{4} \right]_{-2}^{-1}+\left[ \frac{{{x}^{4}}}{4}-3\frac{{{x}^{2}}}{2}+2x \right]_{-1}^{1}+\left[ 3\frac{{{x}^{2}}}{2}-2x \right]_{1}^{2}\]

\[=\left[ 2\times \frac{1}{2}-\frac{1}{4}-2\times \frac{4}{2}+\frac{16}{4} \right]+\left[ \frac{1}{4}-3\times \frac{1}{2}+2(1)-\frac{1}{4}+3\times \frac{1}{2}-2(-1) \right]+\left[ 3\times \frac{4}{2}-2(2)-3\times \frac{1}{2}+2(1) \right]\]

\[=\left[ 1-\frac{1}{4}-4+4 \right]+\left[ \frac{1}{4}-\frac{3}{2}+2-\frac{1}{4}+\frac{3}{2}+2 \right]+\left[ 6-4-3\times \frac{1}{2}+2 \right]\]

\[=\left[ \frac{3}{4} \right]+\left[ 4 \right]+\left[ 4-\frac{3}{2} \right]\]

\[=\frac{3+16+16-6}{4}\]

\[=\frac{29}{4}\]

Thus, \[\int\limits_{-2}^{2}{f(x)dx=\frac{29}{4}}\]


(vii). \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\frac{\pi }{2}}{\frac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\]

Say, \[I=\int\limits_{0}^{\frac{\pi }{2}}{\frac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\to (1)\]

Using – $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

\[I=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left( \frac{\pi }{2}-x \right)\sin \left( \frac{\pi }{2}-x \right)\cos \left( \frac{\pi }{2}-x \right)}{{{\sin }^{4}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{4}}\left( \frac{\pi }{2}-x \right)}dx}\]

Here, $\sin \left( \frac{\pi }{2}-x \right)=\cos x$ and $\cos \left( \frac{\pi }{2}-x \right)=\sin x$

\[\Rightarrow I=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left( \frac{\pi }{2}-x \right)\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}\]

\[=\left[ \int\limits_{0}^{\frac{\pi }{2}}{\frac{\frac{\pi }{2}\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \right]-\left[ \int\limits_{0}^{\frac{\pi }{2}}{\frac{x\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \right]\]

Comparing with equation $(1)$

\[\Rightarrow I=\left[ \int\limits_{0}^{\frac{\pi }{2}}{\frac{\frac{\pi }{2}\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \right]-I\]

\[\Rightarrow 2I=\left[ \frac{\pi }{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \right]\]

\[\Rightarrow I=\frac{\pi }{4}\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}\]

Split the interval – 

\[=\frac{\pi }{4}\left[ \int\limits_{0}^{\frac{\pi }{4}}{\frac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx+}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\frac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \right]\]

Divide the numerator and denominator of the first term by ${{\cos }^{4}}x$ and those of the second term by ${{\sin }^{4}}x$

\[=\frac{\pi }{4}\left[ \int\limits_{0}^{\frac{\pi }{4}}{\frac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}dx+}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\frac{\cot x\cos e{{c}^{2}}x}{{{\cot }^{4}}x+1}dx} \right]\]

Putting – \[{{\tan }^{2}}x=t\Rightarrow 2\tan x{{\sec }^{2}}xdx=dt\]in the first term, and \[{{\cot }^{2}}x=u\Rightarrow -2\cot x{{\operatorname{cosec}}^{2}}xdx=du\]

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=\frac{\pi }{4}\Rightarrow t=1$

  • When $x=\frac{\pi }{4}\Rightarrow u=1$

  • When $x=\frac{\pi }{2}\Rightarrow u=0$

The integral becomes –

\[=\frac{\pi }{4}\left[ \frac{1}{2}\int\limits_{0}^{1}{\frac{dt}{1+{{\operatorname{t}}^{2}}}-\frac{1}{2}}\int\limits_{1}^{0}{\frac{du}{{{u}^{2}}+1}} \right]\]

Interchanging the limits in the second term – 

\[=\frac{\pi }{4}\left[ \frac{1}{2}\int\limits_{0}^{1}{\frac{dt}{1+{{\operatorname{t}}^{2}}}+\frac{1}{2}}\int\limits_{0}^{1}{\frac{du}{{{u}^{2}}+1}} \right]\]

\[=\frac{\pi }{4}\left[ \frac{1}{2}\left( {{\tan }^{-1}}t \right)_{0}^{1}+\frac{1}{2}\left( {{\tan }^{-1}}u \right)_{0}^{1} \right]\]

\[=\frac{\pi }{4}\left[ \frac{1}{2}\left( {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right)+\frac{1}{2}\left( {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right) \right]\]

\[=\frac{\pi }{4}\left[ \frac{1}{2}\left( \frac{\pi }{4}-0 \right)+\frac{1}{2}\left( \frac{\pi }{4}-0 \right) \right]\]

\[=\frac{\pi }{4}\times \frac{\pi }{4}\]

$\therefore I=\frac{{{\pi }^{2}}}{8}$

Thus, $\int\limits_{0}^{\frac{\pi }{2}}{\frac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\frac{{{\pi }^{2}}}{8}$


(viii) \[\int\limits_{0}^{\pi }{\frac{x}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\]

Ans: Given integral – \[\int\limits_{0}^{\pi }{\frac{x}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\]

Say, \[I=\int\limits_{0}^{\pi }{\frac{x}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\to (1)\]

Using – $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$

\[I=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)}{{{a}^{2}}{{\cos }^{2}}\left( \pi -x \right)+{{b}^{2}}{{\sin }^{2}}\left( \pi -x \right)}dx}\]

Here, $\sin (\pi -x)=\sin x\Rightarrow {{\sin }^{2}}(\pi -x)={{\sin }^{2}}x$ and $\cos (\pi -x)=-\cos x\Rightarrow {{\cos }^{2}}(\pi -x)={{\cos }^{2}}x$

\[=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\]

\[=\int\limits_{0}^{\pi }{\frac{\pi }{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}-\int\limits_{0}^{\pi }{\frac{x}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\]

Comparing with equation $(1)$

\[\Rightarrow I=\int\limits_{0}^{\pi }{\frac{\pi }{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}-I\]

\[\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\]

\[\Rightarrow I=\frac{\pi }{2}\int\limits_{0}^{\pi }{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\]

Using – $\int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx},\text{ when f(2a}-x)=f(x)$

\[\Rightarrow I=\frac{\pi }{2}\times 2\int\limits_{0}^{\frac{\pi }{2}}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\]

\[=\pi \int\limits_{0}^{\frac{\pi }{2}}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\]

Split the intervals – 

\[=\pi \left[ \int\limits_{0}^{\frac{\pi }{4}}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}+}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}} \right]\]

Divide the numerator and denominator of the first term by ${{\cos }^{2}}x$ and those of the second term by ${{\sin }^{2}}x$

\[=\pi \left[ \int\limits_{0}^{\frac{\pi }{4}}{\frac{{{\sec }^{2}}xdx}{{{a}^{2}}+{{b}^{2}}{{\tan }^{2}}x}+}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\frac{\cos e{{c}^{2}}xdx}{{{a}^{2}}{{\cot }^{2}}x+{{b}^{2}}}} \right]\]

Putting – \[\tan x=t\Rightarrow {{\sec }^{2}}xdx=dt\]in the first term, and \[\cot x=u\Rightarrow -{{\operatorname{cosec}}^{2}}xdx=du\]

The limits – 

  • When $x=0\Rightarrow t=0$

  • When $x=\frac{\pi }{4}\Rightarrow t=1$

  • When $x=\frac{\pi }{4}\Rightarrow u=1$

  • When $x=\frac{\pi }{2}\Rightarrow u=0$

The integral becomes –

\[=\pi \left[ \int\limits_{0}^{1}{\frac{dt}{{{a}^{2}}+{{b}^{2}}{{\operatorname{t}}^{2}}}-}\int\limits_{1}^{0}{\frac{du}{{{a}^{2}}{{u}^{2}}+{{b}^{2}}}} \right]\]

Interchanging the limits in the second term – 

\[=\pi \left[ \int\limits_{0}^{1}{\frac{dt}{{{a}^{2}}+{{b}^{2}}{{\operatorname{t}}^{2}}}+}\int\limits_{0}^{1}{\frac{du}{{{a}^{2}}{{u}^{2}}+{{b}^{2}}}} \right]\]

\[=\pi \left[ \frac{1}{ab}\left( {{\tan }^{-1}}\frac{bt}{a} \right)_{0}^{1}+\frac{1}{ab}\left( {{\tan }^{-1}}\frac{au}{b} \right)_{0}^{1} \right]\]

\[=\pi \left[ \frac{1}{ab}\left( {{\tan }^{-1}}\frac{b}{a}-{{\tan }^{-1}}0 \right)+\frac{1}{ab}\left( {{\tan }^{-1}}\frac{a}{b}-{{\tan }^{-1}}0 \right) \right]\]

\[=\frac{\pi }{ab}\left[ \left( {{\tan }^{-1}}\frac{b}{a} \right)+\left( {{\tan }^{-1}}\frac{a}{b} \right) \right]\]

\[=\frac{\pi }{ab}\left[ \frac{\pi }{2} \right]\]

\[\Rightarrow I=\frac{{{\pi }^{2}}}{2ab}\]

Thus, \[\int\limits_{0}^{\pi }{\frac{x}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}=\frac{{{\pi }^{2}}}{2ab}\]


47. Evaluate the following integrals

(i) $\int\limits_{\frac{\mathbf{\pi }}{\mathbf{6}}}^{\frac{\mathbf{\pi }}{\mathbf{3}}}{\frac{\mathbf{dx}}{\mathbf{1+}\sqrt{\mathbf{tanx}}}}$

Ans. The given integral can be written as

$\begin{align}

  & I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{dx}{1+\sqrt{\tan x}}} \\ 

 & =\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{dx}{1+\sqrt{\frac{\sin x}{\cos x}}}} \\ 

\end{align}$

$=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\cos x}\text{ }dx}{\sqrt{\cos x}+\sqrt{\sin x}}}$

That is, $I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\cos x}\text{ }dx}{\sqrt{\cos x}+\sqrt{\sin x}}}$                    …… (i)

Now, recall that, since, $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)}dx$.

So,

$I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\cos \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}}{\sqrt{\cos \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}+\sqrt{\sin \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}}}dx$ 

Again, it is known that, $\sin \left( \frac{\pi }{2}-x \right)=\cos x$ and $\cos \left( \frac{\pi }{2}-x \right)=\sin x$.

Therefore,

$I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$                           …… (ii)

Now, adding the equations (i) and (ii), we get

$2I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\cos x}\text{ }dx}{\sqrt{\cos x}+\sqrt{\sin x}}}+\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\text{ }\sqrt{\sin x}\text{ }dx}{\sqrt{\sin x}+\sqrt{\cos x}}}$

$\Rightarrow 2I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\text{ }}dx$

$\Rightarrow 2I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{1\cdot dx}$

$\Rightarrow 2I=\left[ x \right]_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

$\Rightarrow I=\frac{\pi }{12}$

Therefore,

$\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{dx}{1+\sqrt{\tan x}}}=\frac{\pi }{12}$.


(ii) $\int\limits_{\mathbf{0}}^{\mathbf{1}}{\mathbf{si}{{\mathbf{n}}^{\mathbf{-1}}}\left( \frac{\mathbf{2x}}{\mathbf{1+}{{\mathbf{x}}^{\mathbf{2}}}} \right)\mathbf{dx}}$

Ans. Let $x=\tan \theta $.

Then, $dx={{\sec }^{2}}\theta d\theta $

Also, $x\to 0\text{  }\Rightarrow \text{ }\theta \to 0$ and

$x\to 1\text{ }\Rightarrow \text{ }\theta \to \frac{\pi }{4}$.

Therefore, substituting the values, we get

$\int\limits_{\text{0}}^{1}{\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{dx}}=\int\limits_{0}^{\frac{\pi }{4}}{{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)}{{\sec }^{2}}\theta d\theta $

$=\int\limits_{0}^{\frac{\pi }{4}}{{{\sin }^{-1}}\left( \sin 2\theta  \right){{\sec }^{2}}\theta d\theta }$   [since, $\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$]

$=\int\limits_{0}^{\frac{\pi }{4}}{2\theta {{\sec }^{2}}\theta d\theta }$

$=2\int\limits_{0}^{\frac{\pi }{4}}{\theta {{\sec }^{2}}\theta d\theta }$

$=2\left[ \left[ \theta \cdot \tan \theta  \right]_{0}^{\frac{\pi }{4}}-\int\limits_{0}^{\frac{\pi }{4}}{1\cdot \tan \theta d\theta } \right]$       (applying integration by parts)

$=2\left[ \frac{\pi }{4}-\left[ \ln \left| \sec \theta  \right| \right]_{0}^{\frac{\pi }{4}} \right]$

$=\frac{\pi }{2}-2\left[ \ln \left( \sqrt{2} \right)-0 \right]$

$=\frac{\pi }{2}-\ln 2$.

Thus, $\int\limits_{\text{0}}^{1}{\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{dx}}=\frac{\pi }{2}-\ln 2$.


(iii) $\int\limits_{\mathbf{-1}}^{\mathbf{1}}{\mathbf{log}\left( \frac{\mathbf{1+sinx}}{\mathbf{1-sinx}} \right)\mathbf{dx}}$

Ans. Recall that, if $f\left( x \right)$ is an odd function and continuous at the closed interval $\left[ -a,a \right]$, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$.

Let $f\left( x \right)=\log \left( \frac{1+\sin x}{1-\sin x} \right)$, which is continuous on $\left[ -a,a \right]$.

Then, $f\left( -x \right)=\log \left( \frac{1-\sin x}{1+\sin x} \right)$.

Now, $f\left( x \right)+f\left( -x \right)=\log \left( \frac{1+\sin x}{1-\sin x} \right)+\log \left( \frac{1-\sin x}{1+\sin x} \right)$

$=\log \left( \frac{1+\sin x}{1-\sin x}\cdot \frac{1-\sin x}{1+\sin x} \right)$

$=\log \left( 1 \right)$

$=0$.

Therefore, the function $f\left( x \right)=\log \left( \frac{1+\sin x}{1-\sin x} \right)$ is an odd function.

Hence, $\int\limits_{\text{-1}}^{\text{1}}{\text{log}\left( \frac{\text{1+sinx}}{\text{1-sinx}} \right)\text{dx}}=0$.


(iv) $\int\limits_{\mathbf{0}}^{\mathbf{\pi }}{\frac{{{\mathbf{e}}^{\mathbf{cosx}}}}{{{\mathbf{e}}^{\mathbf{cosx}}}\mathbf{+}{{\mathbf{e}}^{\mathbf{-cosx}}}}\mathbf{ }}\mathbf{dx}$

Ans. The given integral can be written as

$I=\int\limits_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\frac{{{\text{e}}^{\text{cosx}}}}{{{\text{e}}^{\text{cosx}}}\text{+}{{\text{e}}^{\text{-cosx}}}}\text{ }}\text{dx}$.                                                            …… (i)

Now, recall that, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Then, $I=\int\limits_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\frac{{{\text{e}}^{\text{cos}\left( \pi -\text{x} \right)}}}{{{\text{e}}^{\text{cos}\left( \pi \text{-x} \right)}}\text{+}{{\text{e}}^{\text{-cos}\left( \pi \text{-x} \right)}}}\text{ }}\text{dx}$

$\Rightarrow I=\int\limits_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\frac{{{\text{e}}^{\text{-cosx}}}}{{{\text{e}}^{\text{-cosx}}}\text{+}{{\text{e}}^{\text{cosx}}}}\text{ }}\text{dx}$    (since $\cos \left( \pi -x \right)=-\cos x$)       …… (ii)    

Add the equations (i) and (ii). Then, we have

$2I=\int\limits_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\frac{{{\text{e}}^{\text{cosx}}}+{{e}^{-\cos x}}}{{{\text{e}}^{\text{cosx}}}\text{+}{{\text{e}}^{\text{-cosx}}}}\text{ }}\text{dx}$

$\Rightarrow 2I=\int\limits_{0}^{\pi }{1\cdot dx}=\left[ x \right]_{0}^{\pi }=\pi $

$\Rightarrow I=\frac{\pi }{2}$

Thus, $\int\limits_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\frac{{{\text{e}}^{\text{cosx}}}}{{{\text{e}}^{\text{cosx}}}\text{+}{{\text{e}}^{\text{-cosx}}}}\text{ }}\text{dx=}\frac{\pi }{2}$.


(v) $\int\limits_{\mathbf{0}}^{\mathbf{\pi }}{\frac{\mathbf{xtanx}}{\mathbf{secx}\cdot \mathbf{cosecx}}}\mathbf{ dx}$

Ans.  The given integral can be written as

$I=\int\limits_{0}^{\pi }{\frac{x\tan x}{\sec x\cdot \cos ecx}}\text{ }dx$.                             …… (1)

Now, recall that, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$.

Then, $I=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)\tan \left( \pi -x \right)}{\sec \left( \pi -x \right)\cdot \cos ec\left( \pi -x \right)}}\text{ }dx$

It is known that, $\tan \left( \pi -x \right)=\tan x$, $\cos ec\left( \pi -x \right)=\cos ecx$, and $\sec \left( \pi -x \right)=\sec x$.

Therefore, $I=\int\limits_{0}^{\pi }{\frac{\left( \pi -x \right)\tan x}{\sec x\cdot \cos ecx}}\text{ }dx$          …… (2)

Adding the equations (1) and (2), gives

$2I=\pi \int\limits_{0}^{\pi }{\frac{\tan x}{\sec x\cdot \cos ecx}}dx$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{{{\sin }^{2}}x}dx$

$\Rightarrow 2I=\frac{2\pi }{2}\int\limits_{0}^{\frac{\pi }{2}}{\left( 1-\cos 2x \right)dx}$

$\Rightarrow 2I=\pi \left[ x-\frac{1}{2}\sin 2x \right]_{0}^{\frac{\pi }{2}}$

$\Rightarrow 2I=\pi \left[ \frac{\pi }{2}-\frac{1}{2}\sin \frac{2\pi }{2}-0 \right]$

$\Rightarrow 2I=\frac{{{\pi }^{2}}}{2}$

$\Rightarrow I=\frac{{{\pi }^{2}}}{4}$.

Thus, $\int\limits_{0}^{\pi }{\frac{x\tan x}{\sec x\cdot \cos ecx}}\text{ }dx=\frac{{{\pi }^{2}}}{4}$.


(vi) $\int\limits_{\mathbf{-a}}^{\mathbf{a}}{\sqrt{\frac{\mathbf{a-x}}{\mathbf{a+x}}}}\mathbf{ dx}$

Ans. Let $x=a\cos \theta $

Then, $dx=-a\sin \theta d\theta $

Also, $x\to -a\text{  }\Rightarrow \theta \to \pi $

and $x\to a\text{ }\Rightarrow \text{ }\theta \to 0$.

Therefore, substituting the values into the given integral, we have

$\int\limits_{-a}^{a}{\sqrt{\frac{a-x}{a+x}}}\text{ }dx=-a\int\limits_{\pi }^{0}{\sqrt{\frac{a-a\cos \theta }{a+a\cos \theta }}}\text{ }\cdot \text{sin}\theta \text{d}\theta $

$=a\int\limits_{0}^{\pi }{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}\text{ }\cdot \text{sin}\theta \text{d}\theta $

$=a\int\limits_{0}^{\pi }{\sqrt{\frac{2{{\sin }^{2}}\frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}}\cdot }\sin \theta d\theta $

$=a\int\limits_{0}^{\pi }{\frac{\sin \frac{\theta }{2}}{\cancel{\cos \frac{\theta }{2}}}\cdot 2\sin \frac{\theta }{2}\cancel{\cos \frac{\theta }{2}}}\text{ }d\theta $

$=a\int\limits_{0}^{\pi }{\left( 1-\cos \theta  \right)d\theta }$    

(since, $1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2}$)

$=a\left[ \theta -\sin \theta  \right]_{0}^{\pi }$

$=a\left[ \pi -\sin \pi -0 \right]$

$=a\pi $.

Thus, \[\int\limits_{\text{-a}}^{\text{a}}{\sqrt{\frac{\text{a-x}}{\text{a+x}}}}\text{ dx}=\text{a}\pi \].


48. $\int\limits_{\mathbf{0}}^{\mathbf{1}}{\left[ \mathbf{2x} \right]\mathbf{dx}}$ where $\left[ {} \right]$ is the greatest integer function.

Ans. First, let $2x=z$.

Then, $2dx=dz$

$\Rightarrow dx=\frac{dz}{2}$.

Also, $x\to 0\text{ }\Rightarrow \text{ z}\to 0$ and

$x\to 1\text{ }\Rightarrow \text{ z}\to \text{2}$.

Therefore, substituting the values we get,

\[\int\limits_{0}^{1}{\left[ 2x \right]dx=}\frac{1}{2}\int\limits_{0}^{2}{\left[ z \right]}dz\]                   …… (i)

Now, note that, by the definition of greatest integer function, 

$\left[ z \right]=\left\{ \begin{align} & 0,\text{ when 0}\le z<1 \\ & 1,\text{ when 1}\le \text{z}<\text{2} \\ \end{align} \right.$

So, from the equation (i), we have

\[\frac{1}{2}\int\limits_{0}^{2}{\left[ z \right]}dz=\frac{1}{2}\left[ \int\limits_{0}^{1}{0dz}+\int\limits_{1}^{2}{1\cdot dz} \right]\]

$=\frac{1}{2}\left[ z \right]_{1}^{2}$

$=\frac{1}{2}\left( 2-1 \right)$

$=\frac{1}{2}$.

Hence, $\int\limits_{\text{0}}^{\text{1}}{\left[ \text{2x} \right]\text{dx}}=\frac{1}{2}$.


49. $\int{{{\mathbf{e}}^{\mathbf{logx+logsinx}}}}\mathbf{dx}$

Ans. The given integral can be written as 

$\int{{{e}^{\log x+\log \sin x}}}dx=\int{{{e}^{\log \left( x\cdot \sin x \right)}}}dx$              [since, $\log \left( a \right)+\log \left( b \right)=\log \left( ab \right)$]

$=\int{{{e}^{{{\log }_{e}}\left( x\sin x \right)}}}dx$

$=\int{x\sin x\text{ }}dx$                                      (since, ${{e}^{{{\log }_{e}}x}}=x$)

$=x\left( -\cos x \right)-\int{1\cdot \left( -\cos x \right)dx}+C$     (applying integration by parts)

$=-x\cos x+\sin x+C$, where $C$ is an integration constant.

Thus, $\int{{{\text{e}}^{\text{logx+logsinx}}}}\text{dx}=-x\cos x+\sin x+C$, where $C$ is an integration constant.


50. $\int{{{\mathbf{e}}^{\mathbf{log}\left( \mathbf{x+1} \right)\mathbf{-logx}}}}\mathbf{dx}$

Ans. The given function can be written as

$\int{{{e}^{\log \left( x+1 \right)-\log x}}}dx=\int{{{e}^{\log \left( \frac{x+1}{x} \right)}}}dx$    (since, $\log \left( a \right)-\log \left( b \right)=\log \left( \frac{a}{b} \right)$)

$=\int{{{e}^{{{\log }_{e}}\left( \frac{x+1}{x} \right)}}}dx$

$=\int{\frac{x+1}{x}dx}$                       (since, ${{e}^{{{\log }_{e}}x}}=x$)

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

$=x+\log x+C$, where $C$ is an integration constant.

Thus, \[\int{{{e}^{\log \left( x+1 \right)-\log x}}}dx=x+\log x+C\], where $C$ is an integration constant.


51. $\int{\frac{\mathbf{sinx}}{\mathbf{sin2x}}\mathbf{dx}}$

Ans. Recall that, $\sin 2x=2\sin x\cos x$.

Then, we have

$\int{\frac{\sin x}{\sin 2x}dx}=\int{\frac{\cancel{\sin x}}{2\cancel{\sin x}\cos x}dx}$

$=\frac{1}{2}\int{\sec xdx}$

$=\frac{1}{2}\log \left| \sec x+\tan x \right|+C$, where $C$ is an integration constant.

$\int{\frac{\sin x}{\sin 2x}dx}=\frac{1}{2}\log \left| \sec x+\tan x \right|+C$, where $C$ is an integration constant.


52. $\int{\mathbf{sinxsin2x}\text{ }\mathbf{dx}}$

Ans. Recall that, $\sin 2x=2\sin x\cos x$.

Then, $\int{\sin x\sin 2x\text{ }dx}=\int{\sin x\cdot \left( 2\sin x\cos x \right)dx}$

$=2\int{{{\sin }^{2}}x\text{ }d\left( \sin x \right)}$

$=\frac{2}{3}{{\sin }^{3}}x+C$, where $C$ is an integration constant.

Hence, $\int{\sin x\sin 2x\text{ }dx}=\frac{2}{3}{{\sin }^{3}}x+C$, where $C$ is an integration constant.


53. $\int\limits_{\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}}^{\frac{\mathbf{\pi }}{\mathbf{4}}}{\left| \mathbf{sinx} \right|\mathbf{dx}}$

Ans. Observe that, the given function is an even function, since $\left| \sin \left( -x \right) \right|=\left| \sin x \right|=\sin x$.

Therefore, the given integral can be written as

$\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left| \sin x \right|dx}=2\int\limits_{0}^{\frac{\pi }{4}}{\left| \sin x \right|dx}$      [$\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$, when $f\left( x \right)$ is even]

Now, since, $0\le x\le \frac{\pi }{4}$, so $\left| \sin x \right|=\sin x$.

Therefore, $\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left| \sin x \right|dx}=2\int\limits_{0}^{\frac{\pi }{4}}{\sin xdx}$

$=-2\left[ \cos x \right]_{0}^{\frac{\pi }{4}}$

$=-2\left( \cos \frac{\pi }{4}-\cos 0 \right)$

$=-2\left( \frac{1}{\sqrt{2}}-1 \right)$

$=2-\sqrt{2}$

Thus, $\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left| \sin x \right|dx}=2-\sqrt{2}$.


54. $\int\limits_{\mathbf{a}}^{\mathbf{b}}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{dx}}\mathbf{+}\int\limits_{\mathbf{b}}^{\mathbf{a}}{\mathbf{f}\left( \mathbf{a+b-x} \right)\mathbf{dx}}$

Ans. The given integral is $\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{b}^{a}{f\left( a+b-x \right)dx}$.

Now, recall that, $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$.

Therefore, we have

$\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{f\left( a+b-x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}+\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$

$=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}-\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$   ( since, $\int\limits_{a}^{b}{f\left( x \right)dx}=-\int\limits_{b}^{a}{f\left( x \right)}$)

$=0$

Hence, $\int\limits_{a}^{b}{f\left( x \right)}+\int\limits_{a}^{b}{f\left( a+b-x \right)dx}=0$.


55. $\int{\frac{\mathbf{1}}{\mathbf{secx+tanx}}}\mathbf{ dx}$

Ans. The given integral can be written as

$\int{\frac{1}{\sec x+\tan x}}\text{ dx}=\int{\frac{dx}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}}$

$=\int{\frac{\cos x\text{ }dx}{1+\sin x}}$

Let $1+\sin x=z$.

Then, $\cos xdx=dz$

Therefore, we have

$\int{\frac{\cos x\text{ }dx}{1+\sin x}}=\int{\frac{dz}{z}}$

$=\log z+C$, where $C$ is an integration constant.

$=\log \left( 1+\sin x \right)+C$.

Thus, $\int{\frac{1}{\sec x+\tan x}}\text{ dx}=\log \left( 1+\sin x \right)+C$, where $C$ is an integration constant.


56. $\int{\frac{\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{x}}{\mathbf{1+cosx}}\mathbf{dx}}$

Ans. The given equation can be written as

$\int{\frac{{{\sin }^{2}}x}{1+\cos x}dx}=\int{\frac{1-{{\cos }^{2}}x}{1+\cos x}dx}$   [since, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$]

$=\int{\frac{\left( 1+\cos x \right)\left( 1-\cos x \right)}{1+\cos x}}\text{ }dx$    [since, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$]

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

$=x-\sin x+C$, where $C$ is an integration constant.

Thus, $\int{\frac{{{\sin }^{2}}x}{1+\cos x}}\text{ }dx=x-\sin x+C$, where $C$ is an integration constant.


57. $\int{\frac{\mathbf{1-tanx}}{\mathbf{1+tanx}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{1-\tan x}{1+\tan x}\text{dx}}=\int{\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\text{ }}dx$

$=\int{\frac{\cos x-\sin x}{\cos x+\sin x}\text{ }}dx$

Now, suppose that, $\cos x+\sin x=z$.

Then, $\left( -\sin x+\cos x \right)dx=dz$

Therefore, substituting the values, we get

$\int{\frac{\cos x-\sin x}{\cos x+\sin x}\text{ }}dx=\int{\frac{dz}{z}}$

$=\log z+C$, where $C$ is an integration constant.

$=\log \left( \cos x+\sin x \right)+C$, substituting the value of $z$.

Hence, $\int{\frac{1-\tan x}{1+\tan x}\text{dx}}=\log \left( \cos x+\sin x \right)+C$, where $C$ is an integration constant.


58. $\int{\frac{{{\mathbf{a}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{x}}}}{{{\mathbf{c}}^{\mathbf{x}}}}\mathbf{ }}\mathbf{dx}$

Ans. The given integral can be written as

$\int{\frac{{{a}^{x}}+{{b}^{x}}}{{{c}^{x}}}\text{ }}dx=\int{\left( \frac{{{a}^{x}}}{{{c}^{x}}}+\frac{{{b}^{x}}}{{{c}^{x}}} \right)\text{ }}dx$

$=\int{{{\left( \frac{a}{c} \right)}^{x}}dx}+\int{{{\left( \frac{b}{c} \right)}^{x}}dx}$

$=\frac{{{\left( \frac{a}{c} \right)}^{x}}}{{{\log }_{e}}\left( \frac{a}{c} \right)}+\frac{{{\left( \frac{b}{c} \right)}^{x}}}{{{\log }_{e}}\left( \frac{b}{c} \right)}+C$, where $C$ is an integration constant.

Thus, $\int{\frac{{{a}^{x}}+{{b}^{x}}}{{{c}^{x}}}\text{ }}dx=\frac{{{\left( \frac{a}{c} \right)}^{x}}}{{{\log }_{e}}\left( \frac{a}{c} \right)}+\frac{{{\left( \frac{b}{c} \right)}^{x}}}{{{\log }_{e}}\left( \frac{b}{c} \right)}+C$, where $C$ is an integration constant.


59. Evaluate

(i) $\int{\frac{\mathbf{si}{{\mathbf{n}}^{\mathbf{-1}}}\sqrt{\mathbf{x}}\mathbf{-co}{{\mathbf{s}}^{\mathbf{-1}}}\sqrt{\mathbf{x}}}{\mathbf{si}{{\mathbf{n}}^{\mathbf{-1}}}\sqrt{\mathbf{x}}\mathbf{+co}{{\mathbf{s}}^{\mathbf{-1}}}\sqrt{\mathbf{x}}}}\mathbf{ dx,  x}\in \left[ \mathbf{0,1} \right]$

Ans. Recall that, ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}$.

Then, the given integral can be written as

$\int{\frac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}}\text{ dx}=\int{\frac{{{\sin }^{-1}}\sqrt{x}-\left( \frac{\pi }{2}-{{\sin }^{-1}}\sqrt{x} \right)}{\frac{\pi }{2}}}$

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

$=\int{\left( \frac{4}{\pi }{{\sin }^{-1}}\sqrt{x}-1 \right)dx}$

$=\frac{4}{\pi }I-x+{{C}_{1}}$ , where \[{{C}_{1}}\] is an integration constant and

$I=\int{{{\sin }^{-1}}\sqrt{x}}dx$.

Let $x={{z}^{2}}$. Then, we get

$dx=2zdz$

Therefore, we have

$I=2\int{{{\sin }^{-1}}z\cdot zdz}$

$=2\left[ {{\sin }^{-1}}\left( z \right)\cdot \frac{{{z}^{2}}}{2}-\int{\frac{1}{\sqrt{1-{{z}^{2}}}}\cdot \frac{{{z}^{2}}}{2}dz} \right]+{{C}_{2}}$    (applying integration by parts)

$={{z}^{2}}{{\sin }^{-1}}z-\int{\frac{{{z}^{2}}}{\sqrt{1-{{z}^{2}}}}dz}+{{C}_{2}}$

$={{z}^{2}}{{\sin }^{-1}}z+\int{\frac{1-{{z}^{2}}-1}{\sqrt{1-{{z}^{2}}}}}\text{ }dz+{{C}_{2}}$

$={{z}^{2}}{{\sin }^{-1}}z+\int{\left( \sqrt{1-{{z}^{2}}}-\frac{1}{\sqrt{1-{{z}^{2}}}} \right)}\text{ }dz+{{C}_{2}}$

Now, recall that, $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}+C$ and

$\int{\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}dx={{\sin }^{-1}}\frac{x}{a}+C$.

Therefore, 

$I={{z}^{2}}{{\sin }^{-1}}z+\frac{z}{2}\sqrt{1-{{z}^{2}}}+\frac{1}{2}{{\sin }^{-1}}z-{{\sin }^{-1}}z+{{C}_{2}}$

$\Rightarrow I={{z}^{2}}{{\sin }^{-1}}z+\frac{z}{2}\sqrt{1-{{z}^{2}}}-\frac{1}{2}{{\sin }^{-1}}z+{{C}_{2}}$

Thus,

 $\int{\frac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}}\text{ dx}=\frac{4}{\pi }I-x+{{C}_{1}}$

$=\frac{4}{\pi }\left( {{z}^{2}}{{\sin }^{-1}}z+\frac{z}{2}\sqrt{1-{{z}^{2}}}-\frac{1}{2}{{\sin }^{-1}}z+{{C}_{2}} \right)-x+{{C}_{2}}$

$=\frac{4}{\pi }\left( x{{\sin }^{-1}}\sqrt{x}+\frac{\sqrt{x-{{x}^{2}}}}{2}-\frac{1}{2}{{\sin }^{-1}}\sqrt{x} \right)-x+C$, substituting the value of $z$.

Hence, $\int{\frac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}}\text{ dx}={{\sin }^{-1}}\sqrt{x}\left[ \frac{2\left( 2x-1 \right)}{\pi } \right]+\frac{2\sqrt{x-{{x}^{2}}}}{\pi }-x+C$, where $C$ is an integration constant.


(ii) $\int{\sqrt{\frac{\mathbf{1-}\sqrt{\mathbf{x}}}{\mathbf{1+}\sqrt{\mathbf{x}}}}}\mathbf{dx}$

Ans. Let $x={{\cos }^{2}}\theta $

Then, $dx=-2\cos \theta \sin \theta $.

Therefore, substituting the values we get,

$\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}}dx=-2\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}\cdot \cos \theta \sin \theta d\theta $

Now, recall that, $1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2}$ and $1+\cos \theta =2{{\cos }^{2}}\frac{\theta }{2}$.

So, we have

$-2\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}\cdot \cos \theta \sin \theta d\theta =-2\int{\frac{\sqrt{2}\sin \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}}\cdot \cos \theta }\left( 2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right)d\theta $

$=-2\int{2{{\sin }^{2}}\frac{\theta }{2}}\cdot \cos \theta d\theta $

$=-2\int{\left( 1-\cos \theta  \right)\cdot \cos \theta d\theta }$

$=-2\int{\left( \cos \theta -{{\cos }^{2}}\theta  \right)d\theta }$

$=-2\sin \theta +\int{\left( 1+\cos 2\theta  \right)}d\theta +C$

$=-2\sin \theta +\theta +\frac{\sin 2\theta }{2}+C$, where $C$ is an integration constant.

$=-2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\frac{2\sqrt{1-x}\cdot \sqrt{x}}{2}+C$

$={{\cos }^{-1}}\sqrt{x}-2\sqrt{1-x}+\sqrt{x-{{x}^{2}}}+C$.

Thus, $\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}}dx={{\cos }^{-1}}\sqrt{x}-2\sqrt{1-x}+\sqrt{x-{{x}^{2}}}+C$, where $C$ is an integration constant.


(iii) $\int{\frac{\sqrt{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1}}\left[ \mathbf{log}\left( {{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1} \right)\mathbf{-2logx} \right]}{{{\mathbf{x}}^{\mathbf{4}}}}\mathbf{ }}\mathbf{dx}$

Ans. The given integral can be written as

$\int{\frac{\sqrt{{{\text{x}}^{\text{2}}}\text{+1}}\left[ \text{log}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{-2logx} \right]}{{{\text{x}}^{\text{4}}}}\text{ }}\text{dx=}\int{\sqrt{1+\frac{1}{{{x}^{2}}}}\left[ \log \left( 1+\frac{1}{{{x}^{2}}} \right) \right]}\cdot \frac{1}{{{x}^{3}}}dx$

Now, suppose that, $1+\frac{1}{{{x}^{2}}}={{z}^{2}}$

Then, $-\frac{2}{{{x}^{3}}}dx=2zdz$

$\Rightarrow \frac{1}{{{x}^{3}}}dx=-zdz$

Therefore, substituting the values we get,

$\int{\sqrt{1+\frac{1}{{{x}^{2}}}}\left[ \log \left( 1+\frac{1}{{{x}^{2}}} \right) \right]}\cdot \frac{1}{{{x}^{3}}}dx=-\int{z\left( 2\log z \right)zdz}$

$=-2\int{\log z\cdot {{z}^{2}}dz}$

$=-2\log z\cdot \frac{{{z}^{3}}}{3}+2\int{\frac{1}{z}\cdot \frac{{{z}^{3}}}{3}dz}$               [applying integration by parts]  

$=-\frac{2}{3}{{z}^{3}}\log z+\frac{2}{9}{{z}^{3}}+C$

$=\frac{2}{3}{{\left( 1+\frac{1}{{{x}^{2}}} \right)}^{3}}\left[ \frac{1}{3}-\log \left( 1+\frac{1}{{{x}^{2}}} \right) \right]+C$, substituting the value of $z$.

Hence, $\int{\frac{\sqrt{{{\text{x}}^{\text{2}}}\text{+1}}\left[ \text{log}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{-2logx} \right]}{{{\text{x}}^{\text{4}}}}\text{ }}\text{dx}=\frac{2}{3}{{\left( 1+\frac{1}{{{x}^{2}}} \right)}^{3}}\left[ \frac{1}{3}-\log \left( 1+\frac{1}{{{x}^{2}}} \right) \right]+C$, where $C$ is an integration constant.


(iv) $\int{\frac{{{\mathbf{x}}^{\mathbf{2}}}}{{{\left( \mathbf{xsinx+cosx} \right)}^{\mathbf{2}}}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}=\int{\frac{x}{\cos x}\cdot \frac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$    …… (i)

Now, note that, $\frac{d}{dx}\left( x\sin x+\cos x \right)=x\cos x+\sin x-\sin x=x\cos x$

Now, applying integration by parts on the equation (i), we have

$\int{\frac{x}{\cos x}\cdot \frac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$

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

$=-\frac{x}{\cos x\left( x\sin x+\cos x \right)}+\int{{{\sec }^{2}}xdx}$

$=\frac{-x}{\cos x\left( x\sin x+\cos x \right)}+\tan x+C$, where $C$ is an integration constant.

$==\frac{-x}{\cos x\left( x\sin x+\cos x \right)}+\frac{\sin x}{\cos x}+C$

$=\frac{-x+\sin x\left( x\sin x+\cos x \right)}{\cos x\left( x\sin x+\cos x \right)}+C$

$=\frac{\sin x-x\cos x}{\cos x+x\sin x}+C$   (simplify)

Thus, 

$\int{\frac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}=\frac{\sin x-x\cos x}{\cos x+x\sin x}+C$, where $C$ is an integration constant.


(v) $\int{\mathbf{si}{{\mathbf{n}}^{\mathbf{-1}}}\sqrt{\frac{\mathbf{x}}{\mathbf{a+x}}}\mathbf{dx}}$

Ans. Let $x=a{{\tan }^{2}}\theta $

Then, $dx=2a\tan \theta {{\sec }^{2}}\theta d\theta $

Therefore, substituting the values we get,

$\int{{{\sin }^{-1}}\sqrt{\frac{x}{a+x}}dx}=\int{{{\sin }^{-1}}\sqrt{\frac{a{{\tan }^{2}}\theta }{a\left( 1+{{\tan }^{2}}\theta  \right)}}}\left( 2a\tan \theta {{\sec }^{2}}\theta  \right)d\theta $

$=2a\int{{{\sin }^{-1}}\left( \frac{\tan \theta }{\sec \theta } \right)\tan \theta {{\sec }^{2}}\theta d\theta }$

$=2a\int{{{\sin }^{-1}}\left( \sin \theta  \right)\tan \theta {{\sec }^{2}}\theta d\theta }$

$=2a\int{\theta \tan \theta {{\sec }^{2}}\theta d\theta }$

$=2a\int{\theta \tan \theta \text{ d}\left( \tan \theta  \right)}$

$=2a\left[ \theta \cdot \frac{{{\tan }^{2}}\theta }{2}-\int{1\cdot \frac{{{\tan }^{2}}\theta }{2}d\theta } \right]$       [applying integration by parts]

$=a\theta {{\tan }^{2}}\theta -a\int{\left( {{\sec }^{2}}\theta -1 \right)d\theta }$

$=a\theta {{\tan }^{2}}\theta -a\tan \theta +a\theta +C$, where $C$ is an integration constant.

$=a{{\tan }^{-1}}\sqrt{\frac{x}{a}}\cdot \frac{x}{a}-a\cdot \sqrt{\frac{x}{a}}+a{{\tan }^{-1}}\sqrt{\frac{x}{a}}+C$, substituting the value of $\theta $.

$=\left( x+a \right){{\tan }^{-1}}\sqrt{\frac{x}{a}}-\sqrt{ax}+C$

Hence, $\int{{{\sin }^{-1}}\sqrt{\frac{x}{a+x}}dx}=\left( x+a \right){{\tan }^{-1}}\sqrt{\frac{x}{a}}-\sqrt{ax}+C$, where $C$ is an integration constant.


(vi) \[\int\limits_{\frac{\mathbf{\pi }}{\mathbf{6}}}^{\frac{\mathbf{\pi }}{\mathbf{3}}}{\frac{\mathbf{sinx+cosx}}{\sqrt{\mathbf{sin2x}}}\mathbf{dx}}\]

Ans. The given integral can be written as

\[\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{1-\left( 1-\sin 2x \right)}}dx}\]

$=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)}}dx}$ [since, $\sin 2x=2\sin x\cos x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$]

$=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}}dx$

Now, note that, $\frac{d}{dx}\left( \sin x-\cos x \right)=\cos x+\sin x$

So, we have

$\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}}dx=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{d\left( \sin x-\cos x \right)}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}}$

$=\left[ {{\sin }^{-1}}\left( \sin x-\cos x \right) \right]_{\frac{\pi }{6}}^{\frac{\pi }{3}}$         [since, $\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}={{\sin }^{-1}}x+C}$]

$={{\sin }^{-1}}\left( \sin \frac{\pi }{3}-\cos \frac{\pi }{3} \right)-{{\sin }^{-1}}\left( \sin \frac{\pi }{6}-\cos \frac{\pi }{6} \right)$

$={{\sin }^{-1}}\left( \frac{\sqrt{3}}{2}-\frac{1}{2} \right)+{{\sin }^{-1}}\left( \frac{\sqrt{3}}{3}-\frac{1}{2} \right)$

$=2{{\sin }^{-1}}\frac{1}{2}\left( \sqrt{3}-1 \right)$.

Hence, $\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}=2{{\sin }^{-1}}\frac{1}{2}\left( \sqrt{3}-1 \right)$.


(vii) $\int\limits_{\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}{\left( \mathbf{sin}\left| \mathbf{x} \right|\mathbf{-cos}\left| \mathbf{x} \right| \right)\mathbf{dx}}$

Ans. Note that, $\left| x \right|=\left\{ \begin{align}

  & -x,\text{ when }-\frac{\pi }{2}\le x\le 0 \\ 

 & x,\text{  when     0}\le \text{x}\le \frac{\pi }{2} \\ 

\end{align} \right.$

Therefore, the given integral can be written as

$\int\limits_{\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\left( \text{sin}\left| \text{x} \right|\text{-cos}\left| \text{x} \right| \right)\text{dx}}=\int\limits_{-\frac{\pi }{2}}^{0}{\left[ \sin \left( -x \right)-\cos \left( -x \right) \right]dx}+\int\limits_{0}^{\frac{\pi }{2}}{\left[ \sin x-\cos x \right]dx}$

$=\int\limits_{-\frac{\pi }{2}}^{0}{\left( -\sin x-\cos x \right)dx}+\left[ -\cos x-\sin x \right]_{0}^{\frac{\pi }{2}}$

$=-\left[ -\cos x+\sin x \right]_{-\frac{\pi }{2}}^{0}-\left[ 0+1-1 \right]$

$=-\left[ -1+0+0+1 \right]-0$

$=0$

Hence, $\int\limits_{\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\left( \text{sin}\left| \text{x} \right|\text{-cos}\left| \text{x} \right| \right)\text{dx}}=0$.


(viii) $\int\limits_{\mathbf{1}}^{\mathbf{2}}{\left[ {{\mathbf{x}}^{\mathbf{2}}} \right]\mathbf{dx}}$, where $\left[ \mathbf{x} \right]$ is greatest integer function.

Ans. Suppose that, ${{x}^{2}}=z$

Then, $2xdx=dz$

$\Rightarrow dx=\frac{1}{2}\frac{dz}{\sqrt{z}}$

Also, $x\to 1\text{ }\Rightarrow \text{ z}\to \text{1}$ and

$x\to 2\text{ }\Rightarrow \text{ z}\to \text{4}$.

Therefore, the given integral can be written as

\[\int\limits_{1}^{2}{\left[ {{x}^{2}} \right]dx}=\frac{1}{2}\int\limits_{1}^{4}{\frac{\left[ z \right]dz}{\sqrt{z}}}\]

$\int\limits_{1}^{2}{\left[ {{x}^{2}} \right]dx}=\frac{1}{2}\left[ \int\limits_{1}^{2}{\frac{\left[ z \right]dz}{\sqrt{z}}}+\int\limits_{2}^{3}{\frac{\left[ z \right]}{\sqrt{z}}dz}+\int\limits_{3}^{4}{\frac{\left[ z \right]}{\sqrt{z}}dz} \right]$               …… (i)

Now, recall the definition of greatest integer function.

$\left[ z \right]=\left\{ \begin{align} & 1,\text{ when 1}\le \text{z}<\text{2} \\ & \text{2, when 2}\le z<3 \\ & 3,\text{ when 3}\le \text{z}<\text{4} \\ \end{align} \right.$

Therefore, from the equation (i) we get,

$\int\limits_{1}^{2}{\left[ {{x}^{2}} \right]dx}=\frac{1}{2}\left[ \int\limits_{1}^{2}{\frac{1\cdot dz}{\sqrt{z}}}+\int\limits_{2}^{3}{\frac{2}{\sqrt{z}}dz}+\int\limits_{3}^{4}{\frac{3}{\sqrt{z}}dz} \right]$

$=\frac{1}{2}\left[ \left[ 2\sqrt{z} \right]_{1}^{2}+4\left[ \sqrt{z} \right]_{2}^{3}+6\left[ \sqrt{z} \right]_{3}^{4} \right]$

$=\frac{1}{2}\left[ 2\sqrt{2}-2+4\sqrt{3}-4\sqrt{2}+12-6\sqrt{3} \right]$

$=\frac{1}{2}\left[ 10-2\sqrt{2}-2\sqrt{3} \right]$

$=5-\sqrt{2}-\sqrt{3}$

Hence, $\int\limits_{1}^{2}{\left[ {{x}^{2}} \right]dx}=5-\sqrt{2}-\sqrt{3}$.


(ix) $\int\limits_{\mathbf{-1}}^{\frac{\mathbf{3}}{\mathbf{2}}}{\left| \mathbf{xsin\pi x} \right|\mathbf{dx}}$

Ans. The given integral can be written as

$\int\limits_{-1}^{\frac{3}{2}}{\left| x\sin \pi x \right|dx}=\int\limits_{-1}^{1}{\left| x\sin \pi x \right|dx}+\int\limits_{1}^{\frac{3}{2}}{\left| x\sin \pi x \right|dx}$                …… (i)

Now, note that $\left| x \right|=-x$ and $\left| \sin \pi x \right|=-\sin \pi x$ for $-1\le x\le 1$.

That is, $\left| x\sin \pi x \right|=\left( -x \right)\left( -\sin \pi x \right)=x\sin \pi x$ when $-1\le x\le 1$.

Also, $\left| x \right|=x$ and $\left| \sin \pi x \right|=-\sin \pi x$ for $1\le x\le \frac{3}{2}$.

That I s, $\left| x\sin \pi x \right|=-x\sin \pi x$ when $1\le x\le \frac{3}{2}$.

Therefore, from equation (i), we have

$\int\limits_{-1}^{\frac{3}{2}}{\left| x\sin \pi x \right|dx}=\int\limits_{-1}^{1}{x\sin \pi xdx}-\int\limits_{1}^{\frac{3}{2}}{x\sin \pi xdx}$                        …… (ii)

Now, $\int{x\sin \pi x}dx=-\frac{x\cos \pi x}{\pi }+\frac{1}{\pi }\int{1\cdot \cos \pi xdx}$

$\int{x\sin \pi x}dx=-\frac{x\cos \pi x}{\pi }+\frac{\sin \pi x}{{{\pi }^{2}}}+C$.

Thus, the equation (ii) becomes

$\int\limits_{-1}^{\frac{3}{2}}{\left| x\sin \pi x \right|dx}=\left[ -\frac{x\cos \pi x}{\pi }+\frac{\sin \pi x}{{{\pi }^{2}}} \right]_{-1}^{1}+\left[ -\frac{x\cos \pi x}{\pi }+\frac{\sin \pi x}{{{\pi }^{2}}} \right]_{1}^{\frac{3}{2}}$

$=\frac{1}{\pi }-\left( -\frac{1}{\pi } \right)-\left[ 0-\frac{1}{{{\pi }^{2}}}-\frac{1}{\pi }-0 \right]$

$=\frac{3}{\pi }+\frac{1}{{{\pi }^{2}}}$

Hence, $\int\limits_{-1}^{\frac{3}{2}}{\left| x\sin \pi x \right|dx}=\frac{3}{\pi }+\frac{1}{{{\pi }^{2}}}$.


Long Answer Type Questions (5 Marks)

60. Evaluate the following integrals:

(i) $\int{\frac{{{\mathbf{x}}^{\mathbf{5}}}\mathbf{+4}}{{{\mathbf{x}}^{\mathbf{5}}}\mathbf{-x}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{{{x}^{5}}+4}{{{x}^{5}}-x}dx}=\int{\left( \frac{{{x}^{5}}}{{{x}^{5}}-x}+\frac{4}{{{x}^{5}}-x} \right)dx}$

$=\int{\frac{{{x}^{5}}}{{{x}^{5}}-x}dx}+\int{\frac{4}{{{x}^{5}}-x}dx}$

$={{I}_{1}}+{{I}_{2}}$ (say)

Now, ${{I}_{1}}=\int{\frac{{{x}^{5}}}{{{x}^{5}}-x}dx}$

$=\int{\frac{{{x}^{5}}dx}{x\left( {{x}^{4}}-1 \right)}}$

$=\int{\frac{{{x}^{4}}}{{{x}^{4}}-1}dx}$

$=\int{\left[ 1-\frac{1}{2\left( {{x}^{2}}+1 \right)}-\frac{1}{4\left( x+1 \right)}+\frac{1}{4\left( x-1 \right)} \right]dx}$     (Taking partial fraction)

$=\int{dx}-\int{\frac{1}{2\left( {{x}^{2}}+1 \right)}dx}-\int{\frac{1}{4\left( x+1 \right)}}dx-\int{\frac{1}{4\left( x-1 \right)}dx}$ (sum rule of integration)

$=x-\frac{1}{2}{{\tan }^{-1}}x-\frac{1}{4}\log \left| x+1 \right|+\frac{1}{4}\log \left| x-1 \right|+{{C}_{1}}$

Also, ${{I}_{2}}=\int{\frac{4}{{{x}^{5}}-x}}dx$

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

$=4\left[ -\int{\frac{dx}{x}}+\int{\frac{d\left( {{x}^{2}}+1 \right)}{4\left( {{x}^{2}}+1 \right)}}+\frac{1}{4}\int{\frac{dx}{x+1}}+\frac{1}{4}\int{\frac{dx}{x-1}} \right]$    (sum rule of integration)

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

Therefore, substituting the values of ${{I}_{1}}$ and ${{I}_{2}}$

 $\begin{align} & \int{\frac{{{x}^{5}}+4}{{{x}^{5}}-x}dx}=x-\frac{1}{2}{{\tan }^{-1}}x-\frac{1}{4}\log \left| x+1 \right|+\frac{1}{4}\log \left| x-1 \right| \\ & \text{ }-4\log \left| x \right|+\log \left| {{x}^{2}}+1 \right|+\log \left| x+1 \right|+\log \left| x-1 \right|+C \\ \end{align}$

$\Rightarrow \int{\frac{{{x}^{5}}+4}{{{x}^{5}}-x}dx}=x-\frac{1}{2}{{\tan }^{-1}}x+\log \left| {{x}^{2}}+1 \right|+\frac{3}{4}\log \left| x+1 \right|+\frac{5}{4}\log \left| x-1 \right|-4\log \left| x \right|+C$, where $C$ is an integration constant.


(ii) $\int{\frac{\mathbf{dx}}{\left( \mathbf{x-1} \right)\left( {{\mathbf{x}}^{\mathbf{2}}}\mathbf{+4} \right)}}$

Ans. The given integral can be written as

$\int{\frac{dx}{\left( x-1 \right)\left( {{x}^{2}}+4 \right)}}=\int{\left[ \frac{1}{5\left( x-1 \right)}-\frac{x}{5\left( {{x}^{2}}+4 \right)}-\frac{1}{5\left( {{x}^{2}}+4 \right)} \right]dx}$  [taking partial fraction]

$\Rightarrow \int{\frac{dx}{\left( x-1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{1}{5}\int{\frac{dx}{x-1}}-\frac{1}{5}\cdot \frac{1}{2}\int{\frac{d\left( {{x}^{2}}+4 \right)}{{{x}^{2}}+4}}-\frac{1}{5}\int{\frac{dx}{{{x}^{2}}+{{2}^{2}}}}$ 

$\Rightarrow \int{\frac{dx}{\left( x-1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{1}{5}\log \left| x-1 \right|-\frac{1}{10}\log \left| {{x}^{2}}+4 \right|-\frac{1}{5}\cdot \frac{1}{2}{{\tan }^{-1}}\frac{x}{2}+C$

$\Rightarrow \int{\frac{dx}{\left( x-1 \right)\left( {{x}^{2}}+4 \right)}}=\frac{1}{5}\log \left| x-1 \right|-\frac{1}{10}\log \left| {{x}^{2}}+4 \right|-\frac{1}{10}{{\tan }^{-1}}\frac{x}{2}+C$, where $C$ is an integration constant.


(iii) $\int{\frac{\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}}{\left( \mathbf{x+1} \right){{\left( \mathbf{x-3} \right)}^{\mathbf{2}}}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{2{{x}^{3}}}{\left( x+1 \right){{\left( x-3 \right)}^{2}}}dx}=\int{\left[ 1-\frac{1}{16\left( x+1 \right)}+\frac{81}{16\left( x-3 \right)}+\frac{27}{4{{\left( x-3 \right)}^{2}}} \right]dx}$ [taking partial fraction]

$\Rightarrow \int{\frac{2{{x}^{3}}}{\left( x+1 \right){{\left( x-3 \right)}^{2}}}dx}=2\left[ \int{dx}-\frac{1}{16}\int{\frac{dx}{x+1}}+\frac{81}{16}\int{\frac{dx}{x-3}}+\frac{27}{4}\int{\frac{dx}{{{\left( x-3 \right)}^{2}}}} \right]$ (by sum rule of integration)

$\Rightarrow \int{\frac{2{{x}^{3}}}{\left( x+1 \right){{\left( x-3 \right)}^{2}}}dx}=2\left[ x-\frac{1}{16}\log \left| x+1 \right|+\frac{81}{16}\log \left| x-3 \right|-\frac{27}{4}\cdot \frac{1}{\left( x-3 \right)} \right]+C$, where $C$ is an integration constant.


(iv) $\int{\frac{{{\mathbf{x}}^{\mathbf{4}}}}{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-16}}\mathbf{dx}}$

Ans. The given integral can be written as

\[\int{\frac{{{x}^{4}}}{{{x}^{4}}-16}dx}=\int{\frac{{{x}^{4}}-16+16}{{{x}^{4}}-16}dx}\]

\[=\int{\left( 1+\frac{16}{{{x}^{4}}-16} \right)dx}\]

$=\int{dx}+16\int{\left[ -\frac{1}{8\left( {{x}^{2}}+4 \right)}-\frac{1}{32\left( x+2 \right)}+\frac{1}{32\left( x-2 \right)} \right]dx}$    (taking partial fraction)

$=x-2\int{\frac{dx}{{{x}^{2}}+{{2}^{2}}}}-\frac{1}{2}\int{\frac{dx}{x+2}}+\frac{1}{2}\int{\frac{dx}{x-2}}+C$, where $C$ is an integration constant.

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

$=x-{{\tan }^{-1}}\frac{x}{2}+\log \sqrt{\frac{x-2}{x+2}}+C$

Hence, $\int{\frac{{{x}^{4}}}{{{x}^{4}}-16}dx}=x-{{\tan }^{-1}}\frac{x}{2}+\log \sqrt{\frac{x-2}{x+2}}+C$, where $C$ is an integration constant.


(v) $\int\limits_{\mathbf{0}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}{\left( \sqrt{\mathbf{tanx}}\mathbf{+}\sqrt{\mathbf{cotx}} \right)\mathbf{dx}}$

Ans. The given integral can be written as

$\int\limits_{0}^{\frac{\pi }{2}}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right)dx}=\int\limits_{0}^{\frac{\pi }{2}}{\left( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right)dx}$

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{\cos x\sin x}}dx}$

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{-\frac{1}{2}\left( -2\sin x\cos x \right)}}dx}$

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{-\frac{1}{2}\left( 1-2\sin x\cos x-1 \right)}}dx}$

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{-\frac{1}{2}\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x-1 \right)}}dx}$

\[=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{-\frac{1}{2}\left[ {{\left( \sin x-\cos x \right)}^{2}}-1 \right]}}dx}\]

\[=\sqrt{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{\left[ 1-{{\left( \sin x-\cos x \right)}^{2}} \right]}}dx}\]

Now, note that, $\frac{d}{dx}\left( \sin x-\cos x \right)=\cos x+\sin x$

Therefore, we have

$\int\limits_{0}^{\frac{\pi }{2}}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right)dx}=\sqrt{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{d\left( \sin x-\cos x \right)}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}}$

$=\sqrt{2}\left[ {{\sin }^{-1}}\left( \sin x-\cos x \right) \right]_{0}^{\frac{\pi }{2}}$

$=\sqrt{2}\left[ {{\sin }^{-1}}\left( \sin \frac{\pi }{2}-\cos \frac{\pi }{2} \right)-{{\sin }^{-1}}\left( \sin 0-\cos 0 \right) \right]$

$=\sqrt{2}\left[ {{\sin }^{-1}}\left( 1-0 \right)-{{\sin }^{-1}}\left( 0-1 \right) \right]$

$=\sqrt{2}\left( \frac{\pi }{2}+\frac{\pi }{2} \right)$

$=\sqrt{2}\pi $

Hence, $\int\limits_{0}^{\frac{\pi }{2}}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right)dx}=\sqrt{2}\pi $.


(vi) $\int{\frac{\mathbf{dx}}{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{+1}}}$

Ans. Note that, $\left( {{x}^{4}}+1 \right)=\left( {{x}^{2}}-\sqrt{2}x+1 \right)\left( {{x}^{2}}+\sqrt{2}x+1 \right)$.

Therefore, by using partial fraction, it can be obtained that, $\frac{1}{{{x}^{4}}+1}=\frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{{{x}^{2}}-\sqrt{2}x+1}+\frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{{{x}^{2}}+\sqrt{2}x+1}$

$\Rightarrow \frac{1}{{{x}^{4}}+1}=\frac{1}{2\sqrt{2}}\frac{-x+\sqrt{2}}{{{x}^{2}}-\sqrt{2}x+1}+\frac{1}{2\sqrt{2}}\cdot \frac{x+\sqrt{2}}{{{x}^{2}}+\sqrt{2}x+1}$

Then, the given integral can be written as

$\int{\frac{\text{dx}}{{{\text{x}}^{\text{4}}}\text{+1}}}=\int{\left[ \frac{1}{2\sqrt{2}}\frac{-x+\sqrt{2}}{{{x}^{2}}-\sqrt{2}x+1}+\frac{1}{2\sqrt{2}}\cdot \frac{x+\sqrt{2}}{{{x}^{2}}+\sqrt{2}x+1} \right]dx}$

$\Rightarrow \int{\frac{\text{dx}}{{{\text{x}}^{\text{4}}}\text{+1}}}=\frac{1}{2\sqrt{2}}\int{\frac{-x+\sqrt{2}}{{{x}^{2}}-\sqrt{2}x+1}dx}+\frac{1}{2\sqrt{2}}\int{\frac{x+\sqrt{2}}{{{x}^{2}}+\sqrt{2}x+1}dx}$

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

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

$+\frac{1}{2}\int{\frac{dx}{{{x}^{2}}-\sqrt{2}x+1}}$

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

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

Hence, 

$\int{\frac{\text{dx}}{{{\text{x}}^{\text{4}}}\text{+1}}}=\frac{1}{4\sqrt{2}}\log \left| \frac{{{x}^{2}}+\sqrt{2}x+1}{{{x}^{2}}-\sqrt{2}x+1} \right|+\frac{1}{2\sqrt{2}}\left[ {{\tan }^{-1}}\left( \sqrt{2}x+1 \right)+{{\tan }^{-1}}\left( \sqrt{2}x-1 \right) \right]+C$

By simplifying, we get

$\int{\frac{\text{dx}}{{{\text{x}}^{\text{4}}}\text{+1}}}=\frac{1}{4\sqrt{2}}\log \left| \frac{{{x}^{2}}+\sqrt{2}x+1}{{{x}^{2}}-\sqrt{2}x+1} \right|+\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)+C$, where $C$ is an integration constant.


(vii) $\int\limits_{\mathbf{0}}^{\infty }{\frac{\mathbf{xta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}}{{{\left( \mathbf{1+}{{\mathbf{x}}^{\mathbf{2}}} \right)}^{\mathbf{2}}}}\mathbf{dx}}$

Ans. Suppose that, $x=\tan z$

Then, $dx={{\sec }^{2}}zdz$

Also, $x\to 0$ $\Rightarrow \text{ z}\to \text{0}$ and

$x\to \infty $ $\Rightarrow \text{ z}\to \frac{\pi }{2}$.

Therefore, substituting the values we get,

$\int\limits_{0}^{\infty }{\frac{x{{\tan }^{-1}}x}{{{\left( 1+{{x}^{2}} \right)}^{2}}}dx}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{z\tan z\cdot {{\sec }^{2}}z}{{{\left( 1+{{\tan }^{2}}z \right)}^{2}}}dz}$

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{z\tan z}{{{\sec }^{2}}z}dz}$              (${{\sec }^{2}}z-{{\tan }^{2}}z=1$)

$=\int\limits_{0}^{\frac{\pi }{2}}{\frac{z\sin z\cdot {{\cos }^{2}}z}{\cos z}dz}$

$=\int\limits_{0}^{\frac{\pi }{2}}{z\sin z\cos z\text{ }dz}$

$=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{z\sin 2z\text{ }dz}$

$=\frac{1}{2}\left[ \left[ z\cdot \left( -\frac{\cos 2z}{2} \right) \right]_{0}^{\frac{\pi }{2}}+\int\limits_{0}^{\frac{\pi }{2}}{1\cdot \frac{\cos 2z}{2}dz} \right]$           (applying integration by parts)

$=\frac{1}{2}\left[ -\frac{\pi }{4}\cos \pi +\frac{1}{2}\left[ \frac{\sin 2\pi }{2} \right]_{0}^{\frac{\pi }{2}} \right]$

$=\frac{1}{2}\left( \frac{\pi }{2}+0 \right)$

$=\frac{\pi }{8}$

Hence, $\int\limits_{0}^{\infty }{\frac{x{{\tan }^{-1}}x}{{{\left( 1+{{x}^{2}} \right)}^{2}}}dx}=\frac{\pi }{8}$


61. Evaluate the following integrals as limit of sums:

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

Ans. Recall the definition of definite integral as limit of sums:

$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+\cdot \cdot \cdot +f\left( a+\left( n-1 \right)h \right) \right]$

and $h=\frac{b-a}{n}$.

Now, in the given integral, $a=2$, $b=4$, $f\left( x \right)=2x+1$.

Therefore, $h=\frac{4-2}{n}=\frac{2}{n}$

So, the given integral can be expressed as

$\int\limits_{\text{2}}^{\text{4}}{\left( \text{2x+1} \right)\text{dx}}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 2 \right)+f\left( 2+h \right)+f\left( 2+2h \right)+\cdot \cdot \cdot +f\left( 2+\left( n-1 \right)h \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left. \left[ \left\{ 2\left( 2 \right)+1 \right\}+ \right.\left\{ 2\left( 2+h \right)+1 \right\}+\left\{ 2\left( 2+2h \right) \right\}+1+\cdot \cdot \cdot +\left\{ 2\left( 2+\left( n-1 \right)h \right)+1 \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5+\left( 5+h \right)+\left( 5+2h \right)+\cdot \cdot \cdot +\left( 5+\left( n-1 \right)h \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5n+h\left( 1+2+3+\cdot \cdot \cdot +\left( n-1 \right) \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5n+h\cdot \frac{n\left( n-1 \right)}{2} \right]$         $\left[ \text{since, }\sum\limits_{i=1}^{n-1}{i}=1+2+\cdot \cdot \cdot +\left( n-1 \right)=\frac{n\left( n-1 \right)}{2} \right]$

Now, since, $h\to 0$ and $h=\frac{2}{n}$ $\Rightarrow n\to \infty $.

Therefore,

$\int\limits_{\text{2}}^{\text{4}}{\left( \text{2x+1} \right)\text{dx}}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 5n\cdot \frac{2}{n}+\frac{{{2}^{2}}}{{{n}^{2}}}\cdot \frac{n\left( n-1 \right)}{2} \right]$

$\Rightarrow \int\limits_{\text{2}}^{\text{4}}{\left( \text{2x+1} \right)\text{dx}}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 10+\frac{4\left( n-1 \right)}{n} \right]=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 10+4\left( 1-\frac{1}{n} \right) \right]$

$\Rightarrow \int\limits_{2}^{4}{\left( 2x+1 \right)dx}=14$


(ii) $\int\limits_{\mathbf{0}}^{\mathbf{2}}{\left( {{\mathbf{x}}^{\mathbf{2}}}\mathbf{+3} \right)\mathbf{dx}}$

Ans. Recall the definition of definite integral as limit of sums:

$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+\cdot \cdot \cdot +f\left( a+\left( n-1 \right)h \right) \right]$

and $h=\frac{b-a}{n}$.

Now, in the given integral, $a=0$, $b=2$, $f\left( x \right)={{x}^{2}}+3$.

Therefore, $h=\frac{2-0}{n}=\frac{2}{n}$

So, the given integral can be expressed as

$\int\limits_{0}^{2}{\left( {{x}^{2}}+3 \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 0 \right)+f\left( 0+h \right)+f\left( 0+2h \right)+\cdot \cdot \cdot +f\left( 0+\left( n-1 \right)h \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3+\left\{ {{h}^{2}}+3 \right\}+\left\{ {{\left( 2h \right)}^{2}}+3 \right\}+\cdot \cdot \cdot +\left\{ {{\left( \left( n-1 \right)h \right)}^{2}}+3 \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3n+{{h}^{2}}\left\{ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{\left( n-1 \right)}^{2}} \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3n+{{h}^{2}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$   $\left[ \because \text{ }{{1}^{2}}+{{2}^{2}}+\cdot \cdot \cdot +{{n}^{2}}=\frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$

Now, since, $h\to 0$ and $h=\frac{2}{n}$, so $n\to \infty $.

Therefore, we have

 $\int\limits_{0}^{2}{\left( {{x}^{2}}+3 \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 3n\cdot \frac{2}{n}+{{\left( \frac{2}{n} \right)}^{3}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 6+\frac{8}{6}\left( 1-\frac{1}{n} \right)\left( 2-\frac{1}{n} \right) \right]$

$=6+\frac{4}{3}\left( 2 \right)$

Thus,

$\int\limits_{0}^{2}{\left( {{x}^{2}}+3 \right)dx}=\frac{26}{3}$.


(iii) $\int\limits_{\mathbf{1}}^{\mathbf{3}}{\left( \mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-2x+4} \right)\mathbf{dx}}$

Ans. Recall the definition of definite integral as limit of sums:

$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+\cdot \cdot \cdot +f\left( a+\left( n-1 \right)h \right) \right]$

and $h=\frac{b-a}{n}$.

Now, in the given integral, $a=1$, $b=3$, $f\left( x \right)=3{{x}^{2}}-2x+4$.

Therefore, $h=\frac{3-1}{n}=\frac{2}{n}$

So, the given integral can be expressed as

$\int\limits_{1}^{3}{\left( 3{{x}^{2}}-2x+4 \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 1 \right)+f\left( 1+h \right)+f\left( 1+2h \right)+\cdot \cdot \cdot +f\left( 1+\left( n-1 \right)h \right) \right]$

\[=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \left\{ 3{{\left( 1 \right)}^{2}}-2\left( 1 \right)+4 \right\}+\left\{ 3{{\left( 1+h \right)}^{2}}-2\left( 1+h \right)+4 \right\} \right.\]

\[\left. +\left\{ 3{{\left( 1+2h \right)}^{2}}-2\left( 1+2h \right)+4 \right\}+\cdot \cdot \cdot +\left\{ 3{{\left( 1+\left( n-1 \right)h \right)}^{2}}-2\left( 1+\left( n-1 \right)h \right)+4 \right\} \right]\]

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5+\left( 5+4h+3{{h}^{2}} \right)+\left( 5+8h+12{{h}^{2}} \right)+\cdot \cdot \cdot +\left( 5+4\left( n-1 \right)h+3{{\left( n-1 \right)}^{2}}{{h}^{2}} \right) \right]$

\[=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5n+\left\{ 4h+8h+\cdot \cdot \cdot +4\left( n-1 \right)h \right\}+\left\{ 3{{h}^{2}}+12{{h}^{2}}+\cdot \cdot \cdot +3{{\left( n-1 \right)}^{2}}{{h}^{2}} \right\} \right]\]

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5n+4h\left\{ 1+2+\cdot \cdot \cdot +\left( n-1 \right) \right\}+3{{h}^{2}}\left\{ {{1}^{2}}+{{2}^{2}}+\cdot \cdot \cdot +{{\left( n-1 \right)}^{2}} \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 5n+4h\cdot \frac{n\left( n-1 \right)}{2}+3{{h}^{2}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$ 

Now, since, $h\to 0$ and $h=\frac{2}{n}$, so $n\to \infty $.

Therefore, we have

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 5n\cdot \frac{2}{n}+4\cdot \frac{{{2}^{2}}}{{{n}^{2}}}\cdot \frac{n\left( n-1 \right)}{2}+3\cdot \frac{{{2}^{3}}}{{{n}^{3}}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 10+8\cdot \left( 1-\frac{1}{n} \right)+4\left( 1-\frac{1}{n} \right)\left( 2-\frac{1}{n} \right) \right]$

$\begin{align}

  & =10+8+8 \\ 

 & =26 \\ 

\end{align}$

Hence, $\int\limits_{1}^{3}{\left( 3{{x}^{2}}-2x+4 \right)dx}=26$


(iv) $\int\limits_{\mathbf{0}}^{\mathbf{4}}{\left( \mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{e}}^{\mathbf{2x}}} \right)\mathbf{dx}}$

Ans. Recall the definition of definite integral as limit of sums:

$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+\cdot \cdot \cdot +f\left( a+\left( n-1 \right)h \right) \right]$

and $h=\frac{b-a}{n}$.

Now, in the given integral, $a=0$, $b=4$, $f\left( x \right)=3{{x}^{2}}+{{e}^{2x}}$.

Therefore, $h=\frac{4-0}{n}=\frac{4}{n}$

So, the given integral can be expressed as

$\int\limits_{0}^{4}{\left( 3{{x}^{2}}+{{e}^{2x}} \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 0 \right)+f\left( 0+h \right)+f\left( 0+2h \right)+\cdot \cdot \cdot +f\left( 0+\left( n-1 \right)h \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3\cdot {{0}^{2}}+{{e}^{2\times 0}}+\left( 3{{h}^{2}}+{{e}^{2h}} \right)+\left\{ 3{{\left( 2h \right)}^{2}}+{{e}^{2\left( 2h \right)}} \right\}+\cdot \cdot \cdot +\left\{ 3{{\left( n-1 \right)}^{2}}{{h}^{2}}+{{e}^{2\left( n-1 \right)h}} \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 1+3{{h}^{2}}+{{e}^{2h}}+12{{h}^{2}}+{{e}^{4h}}+\cdot \cdot \cdot +3{{\left( n-1 \right)}^{2}}{{h}^{2}}+{{e}^{2\left( n-1 \right)h}} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3{{h}^{2}}\left\{ {{1}^{2}}+{{2}^{2}}+\cdot \cdot \cdot +{{\left( n-1 \right)}^{2}} \right\}+\left\{ 1+{{e}^{2h}}+{{e}^{4h}}+\cdot \cdot \cdot +{{e}^{2\left( n-1 \right)h}} \right\} \right]$

Now, observe that, the numbers $1,{{e}^{2h}},{{e}^{4h}},...,{{e}^{2\left( n-1 \right)h}}$ forms a G.P series with the common ratio $r=\frac{{{e}^{2h}}}{1}={{e}^{2h}}$.

Therefore, $S=1+{{e}^{2h}}+{{e}^{4h}}+\cdot \cdot \cdot +{{e}^{2\left( n-1 \right)h}}=1\left\{ \frac{{{\left( {{e}^{2h}} \right)}^{n}}-1}{{{e}^{2h}}-1} \right\}=\frac{{{e}^{2nh}}-1}{{{e}^{2h}}-1}$

Thus, $\int\limits_{0}^{4}{\left( 3{{x}^{2}}+{{e}^{2x}} \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3{{h}^{2}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6}+\frac{{{e}^{2nh}}-1}{{{e}^{2h}}-1} \right]$

Now, since, $h\to 0$ and $h=\frac{4}{n}$, so $n\to \infty $.

Therefore, we have

$\int\limits_{0}^{4}{\left( 3{{x}^{2}}+{{e}^{2x}} \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 3\cdot \frac{{{4}^{3}}}{{{n}^{3}}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]+\underset{n\to \infty }{\mathop{\lim }}\,\frac{4}{n}\left[ \frac{{{e}^{2nh}}-1}{2h\cdot \frac{{{e}^{2h}}-1}{2h}} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 32\left( 1-\frac{1}{n} \right)\left( 2-\frac{1}{n} \right) \right]+4\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2nh-1}{2nh}\cdot \frac{1}{\frac{{{e}^{2h}}-1}{2h}} \right)$

$=64+4\underset{n\to \infty }{\mathop{\lim }}\,\frac{2nh-1}{2nh}\cdot \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{\frac{{{e}^{2h}}-1}{2h}}$

$=64+4\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{e}^{2nh}}-1}{2nh}\cdot 1$    [since, $\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{h}}-1}{h}=1$]

$=64+4\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{e}^{2n\cdot \frac{4}{n}}}-1}{2n\left( \frac{4}{n} \right)}$

$=64+4\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{e}^{8}}-1}{8}$

$=64+\frac{{{e}^{8}}-1}{2}$.

Hence, $\int\limits_{0}^{4}{\left( 3{{x}^{2}}+{{e}^{2x}} \right)dx}=64+\frac{{{e}^{8}}-1}{2}$.


(v) $\int\limits_{\mathbf{2}}^{\mathbf{5}}{\left( {{\mathbf{x}}^{\mathbf{2}}}\mathbf{+3x} \right)\mathbf{dx}}$

Ans. Recall the definition of definite integral as limit of sums:

$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+\cdot \cdot \cdot +f\left( a+\left( n-1 \right)h \right) \right]$

and $h=\frac{b-a}{n}$.

Now, in the given integral, $a=2$, $b=5$, $f\left( x \right)={{x}^{2}}+3x$.

Therefore, $h=\frac{5-2}{n}=\frac{3}{n}$

So, the given integral can be expressed as

$\int\limits_{2}^{5}{\left( {{x}^{2}}+3x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 2 \right)+f\left( 2+h \right)+f\left( 2+2h \right)+\cdot \cdot \cdot +f\left( 2+\left( n-1 \right)h \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \left\{ {{2}^{2}}+3\left( 2 \right) \right\}+\left\{ {{\left( 2+h \right)}^{2}}+3\left( 2+h \right) \right\}+\left\{ {{\left( 2+2h \right)}^{2}}+3\left( 2+2h \right) \right\} \right.$

$\left. +\cdot \cdot \cdot +\left\{ {{\left( 2+\left( n-1 \right) \right)}^{2}}+3\left( 2+\left( n-1 \right)h \right) \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 10+\left( 10+7h+{{h}^{2}} \right)+\left( 10+14h+4{{h}^{2}} \right)+\cdot \cdot \cdot +\left( 10+7\left( n-1 \right)h+4\left( n-1 \right){{h}^{2}} \right) \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 10n+7h\left\{ 1+2+3+\cdot \cdot \cdot +\left( n-1 \right) \right\}+{{h}^{2}}\left\{ {{1}^{2}}+{{2}^{2}}+\cdot \cdot \cdot +{{\left( n-1 \right)}^{2}} \right\} \right]$

It is known that, $1+2+3+\cdot \cdot \cdot +\left( n-1 \right)=\frac{n\left( n-1 \right)}{2}$ and 

${{1}^{2}}+{{2}^{2}}+\cdot \cdot \cdot +{{\left( n-1 \right)}^{2}}=\frac{n\left( n-1 \right)\left( 2n-1 \right)}{6}$.

Therefore, we have

$\int\limits_{2}^{5}{\left( {{x}^{2}}+3x \right)dx}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 10n+7h\cdot \frac{n\left( n-1 \right)}{2}+{{h}^{2}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$

Now, since, $h\to 0$ and $h=\frac{3}{n}$, so $n\to \infty $.

Therefore, we have

$\int\limits_{2}^{5}{\left( {{x}^{2}}+3x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 10n\cdot \frac{3}{n}+7\cdot \frac{{{3}^{2}}}{{{n}^{2}}}\cdot \frac{n\left( n-1 \right)}{2}+\frac{{{3}^{3}}}{{{n}^{3}}}\cdot \frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$

$\Rightarrow \int\limits_{2}^{5}{\left( {{x}^{2}}+3x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 30+\frac{63}{2}\cdot \left( 1-\frac{1}{n} \right)+\frac{9}{2}\left( 1-\frac{1}{n} \right)\left( 2-\frac{1}{n} \right) \right]$

$\Rightarrow \int\limits_{2}^{5}{\left( {{x}^{2}}+3x \right)dx}=30+\frac{63}{2}+9=\frac{141}{2}$.


62. Evaluate

(i) $\int\limits_{\mathbf{0}}^{\mathbf{1}}{\mathbf{co}{{\mathbf{t}}^{\mathbf{-1}}}\left( \mathbf{1-x+}{{\mathbf{x}}^{\mathbf{2}}} \right)\mathbf{dx}}$

Ans. The given integral can be written as

$\int\limits_{0}^{1}{{{\cot }^{-1}}\left( 1-x+{{x}^{2}} \right)dx}=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \frac{1}{1-x+{{x}^{2}}} \right)dx}$

$=\int\limits_{0}^{1}{{{\tan }^{-1}}\left[ \frac{x+1-x}{1-x\left( 1-x \right)} \right]dx}$

$=\int\limits_{0}^{1}{\left[ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( 1-x \right) \right]dx}$    [since, ${{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)={{\tan }^{-1}}a+{{\tan }^{-1}}b$]

$=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( x \right)dx}+\int\limits_{0}^{1}{{{\tan }^{-1}}\left( 1-x \right)dx}$

$=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( x \right)dx}+\int\limits_{0}^{1}{{{\tan }^{-1}}\left\{ 1-\left( 1-x \right) \right\}dx}$   (since $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$)

$=2\int\limits_{0}^{1}{{{\tan }^{-1}}\left( x \right)dx}$

$=2\int\limits_{0}^{1}{{{\tan }^{-1}}\left( x \right)\cdot 1dx}$

$=2\left[ \left[ {{\tan }^{-1}}\left( x \right)\cdot x \right]_{0}^{1}-\int\limits_{0}^{1}{\frac{x}{1+{{x}^{2}}}dx} \right]$

$=2\left[ \left[ {{\tan }^{-1}}\left( x \right)\cdot x \right]_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{\frac{2x}{1+{{x}^{2}}}dx} \right]$

$=2\left[ \left[ {{\tan }^{-1}}\left( x \right)\cdot x \right]_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{\frac{d\left( 1+{{x}^{2}} \right)}{1+{{x}^{2}}}dx} \right]$

$=2\left[ \frac{\pi }{4}-\left[ \frac{1}{2}\log \left| 1+{{x}^{2}} \right| \right]_{0}^{1} \right]$

$=\frac{\pi }{2}-\left( \log 2-\log 1 \right)$

$=\frac{\pi }{2}-\log 2$.

Hence, $\int\limits_{0}^{1}{{{\cot }^{-1}}\left( 1-x+{{x}^{2}} \right)dx}=\frac{\pi }{2}-\log 2$.


(ii) $\int{\frac{\mathbf{dx}}{\left( \mathbf{sinx-2cosx} \right)\left( \mathbf{2sinx+cosx} \right)}}$

Ans. The given integral can be written as

$\int{\frac{dx}{\left( \sin x-2\cos x \right)\left( 2\sin x+\cos x \right)}}=\int{\frac{dx}{2{{\sin }^{2}}x+\sin x\cos x-4\sin x\cos x-2{{\cos }^{2}}x}}$

$=\int{\frac{dx}{2{{\sin }^{2}}x-3\sin x\cos x-2{{\cos }^{2}}x}}$

$=\int{\frac{{{\sec }^{2}}x\text{ }dx}{2{{\tan }^{2}}x-3\tan x-2}}$     (divide the numerator and denominator by ${{\cos }^{2}}x$)

$=\frac{1}{2}\int{\frac{d\left( \tan x \right)}{{{\tan }^{2}}x-\frac{3}{2}\tan x-1}}$        [since, $\frac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$]

$=\frac{1}{2}\int{\frac{d\left( \tan x \right)}{{{\left( \tan x-\frac{3}{4} \right)}^{2}}-{{\left( \frac{5}{4} \right)}^{2}}}}$ 

$=\frac{1}{2}\times \frac{1}{2\left( \frac{5}{4} \right)}\log \left| \frac{\tan x-\frac{3}{4}-\frac{5}{4}}{\tan x-\frac{3}{4}+\frac{5}{4}} \right|+C$      (since, $\int{\frac{dx}{{{x}^{2}}-{{a}^{2}}}=\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|+C}$)

$=\frac{1}{5}\log \left| \frac{2\tan x-4}{2\tan x+1} \right|+C$, where $C$ is an integration constant.

Hence, $\int{\frac{dx}{\left( \sin x-2\cos x \right)\left( 2\sin x+\cos x \right)}}=\frac{1}{5}\log \left| \frac{2\tan x-4}{2\tan x+1} \right|+C$, where $C$ is an integration constant.


(iii) $\int\limits_{\mathbf{0}}^{\mathbf{1}}{\frac{\mathbf{log}\left( \mathbf{1+x} \right)}{\mathbf{1+}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{dx}}$

Ans. Suppose that, $x=\tan \theta $

Then, $dx={{\sec }^{2}}\theta d\theta $

Also, $x\to 0$ $\Rightarrow \text{ }\theta \to \text{0}$ and

$x\to 1\text{ }\Rightarrow \text{ }\theta \to \frac{\pi }{4}$

Therefore, substituting the values in the given integral we get,

$\int\limits_{0}^{1}{\frac{\log \left( 1+x \right)}{1+{{x}^{2}}}dx}=\int\limits_{0}^{\frac{\pi }{4}}{\frac{\log \left( 1+\tan \theta  \right){{\sec }^{2}}\theta \text{ d}\theta }{1+{{\tan }^{2}}\theta }}$

$=\int\limits_{0}^{\frac{\pi }{4}}{\frac{\log \left( 1+\tan \theta  \right)\cancel{{{\sec }^{2}}\theta }\text{ d}\theta }{\cancel{{{\sec }^{2}}\theta }}}$    (since, ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$)

$=\int\limits_{0}^{\frac{\pi }{4}}{\log \left( 1+\tan \theta  \right)d\theta }=I$ (say)

Now, since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$, so we have

$I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left[ 1+\tan \left( \frac{\pi }{4}-\theta  \right) \right]d\theta }$

$\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left[ 1+\frac{\tan \left( \frac{\pi }{4} \right)-\tan \theta }{1+\tan \left( \frac{\pi }{4} \right)\tan \theta } \right]d\theta }$     (since, $\tan \left( x-y \right)=\frac{\tan x-\tan y}{1+\tan x\tan y}$)

$\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left[ 1+\frac{1-\tan \theta }{1+\tan \theta } \right]d\theta }$

$\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\log \left( \frac{2}{1+\tan \theta } \right)d\theta }$

$\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\left[ \log 2-\log \left( 1+\tan \theta  \right) \right]d\theta }$

$\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{\log 2d\theta }-\int\limits_{0}^{\frac{\pi }{4}}{\log \left( 1+\tan \theta  \right)d\theta }$

$\Rightarrow I=\log 2\left[ \theta  \right]_{0}^{\frac{\pi }{4}}-I$

$\Rightarrow 2I=\frac{\pi }{4}\log 2$

$\Rightarrow I=\frac{\pi }{8}\log 2$

Hence, $\int\limits_{0}^{1}{\frac{\log \left( 1+x \right)}{1+{{x}^{2}}}dx}=\frac{\pi }{8}\log 2$.


(iv) \[\int\limits_{\mathbf{0}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}{\left( \mathbf{2logsinx-logsin2x} \right)\mathbf{dx}}\]

Ans. Recall that, $\sin 2x=2\sin x\cos x$ and $\log \left( ab \right)=\log a+\log b$.

Then, the given integral can be written as

\[I=\int\limits_{0}^{\frac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}\]

\[=\int\limits_{0}^{\frac{\pi }{2}}{\left[ 2\log \sin x-\log \left( 2\sin x\cos x \right) \right]dx}\]

$=\int\limits_{0}^{\frac{\pi }{2}}{\left[ 2\log \sin x-\log 2-\log \sin x-\log \cos x \right]dx}$

\[=\int\limits_{0}^{\frac{\pi }{2}}{\left( \log \sin x \right)dx}-\int\limits_{0}^{\frac{\pi }{2}}{\log 2dx}-\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \cos x \right)dx}\] 

$=\int\limits_{0}^{\frac{\pi }{2}}{\log \left[ \sin \left( \frac{\pi }{2}-x \right) \right]}-\log 2\left[ x \right]_{0}^{\frac{\pi }{2}}-\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \cos x \right)dx}$    [$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$]

$=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \cos x \right)}-\frac{\pi }{2}\log 2-\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \cos x \right)}$

$\Rightarrow I=-\frac{\pi }{2}\log 2=\frac{\pi }{2}\log \frac{1}{2}$

$\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}=\frac{\pi }{2}\log \frac{1}{2}$.


63. $\int{\frac{\mathbf{1}}{\mathbf{sinx+sin2x}}\mathbf{dx}}$

Ans. The given integral can be written as

$\int{\frac{1}{\sin x+\sin 2x}dx}=\int{\frac{dx}{\sin x+2\sin x\cos x}}$

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

$=\int{\frac{\sin xdx}{{{\sin }^{2}}x\left( 1+2\cos x \right)}}$  (multiply numerator and denominator by $\sin x$)

$=\int{\frac{\sin xdx}{\left( 1-{{\cos }^{2}}x \right)\left( 1+2\cos x \right)}}$

Substitute $\cos x=z$

Then, $-\sin xdx=dz$

$=-\int{\frac{dz}{\left( 1-{{z}^{2}} \right)\left( 1+2z \right)}}$

$=-\int{\left( -\frac{1}{6}\cdot \frac{1}{z-1}-\frac{1}{2}\cdot \frac{1}{z+1}+\frac{4}{3}\cdot \frac{1}{2z+1} \right)dz}$

$=\frac{1}{6}\int{\frac{dz}{z-1}}-\frac{1}{2}\int{\frac{dz}{z+1}}+\frac{4}{3}\int{\frac{dz}{2z+1}}$

$=\frac{1}{6}\log \left| z-1 \right|-\frac{1}{2}\log \left| z+1 \right|+\frac{4}{3}\cdot \frac{1}{2}\log \left| 2z+1 \right|+C$, where $C$ is an integration constant.

Hence, by the substituting the value of $z$ we get, $\int{\frac{1}{\sin x+\sin 2x}dx}=\frac{1}{6}\log \left| \cos x-1 \right|-\frac{1}{2}\log \left| \cos x+1 \right|+\frac{2}{3}\log \left| 2\cos x+1 \right|+C$, where $C$ is an integration constant.


64. $\int{\frac{\left( \mathbf{3sin\theta -2} \right)\mathbf{cos\theta }}{\mathbf{5-co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{\theta -4sin\theta }}\mathbf{d\theta }}$

Ans. The given integral can be written as

$\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{5-{{\cos }^{2}}\theta -4\sin \theta }d\theta }=\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{5-\left( 1-{{\sin }^{2}}\theta  \right)-4\sin \theta }d\theta }$

$=\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{4+{{\sin }^{2}}\theta -4\sin \theta }d\theta }$

$=\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{{{\left( 2-\sin \theta  \right)}^{2}}}}$

Suppose that, $2-\sin \theta =z$

Then, $-\cos \theta d\theta =dz$

Therefore, substituting the values the integral becomes

$\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{5-{{\cos }^{2}}\theta -4\sin \theta }d\theta }=-\int{\frac{\left[ 3\left( 2-z \right)-2 \right]dz}{{{z}^{2}}}}$

$=-4\int{\frac{\left( 4-3z \right)dz}{{{z}^{2}}}}$

$=-4\int{\frac{dz}{{{z}^{2}}}+3\int{\frac{zdz}{{{z}^{2}}}}}$

$=\frac{4}{z}+3\log \left| z \right|+C$

Thus, substituting the value of $z$, we get

$\int{\frac{\left( 3\sin \theta -2 \right)\cos \theta }{5-{{\cos }^{2}}\theta -4\sin \theta }d\theta }=\frac{4}{2-\sin \theta }+3\log \left| 2-\sin \theta  \right|+C$, where $C$ is an integration constant.


65. $\int{\mathbf{se}{{\mathbf{c}}^{\mathbf{3}}}\mathbf{xdx}}$

Ans. The given integral can be written as

$I=\int{{{\sec }^{3}}xdx}$

$=\int{\sec x\cdot {{\sec }^{2}}x\text{ }dx}$

$=\sec x\int{{{\sec }^{2}}xdx}-\int{\left\{ \frac{d}{dx}\left( \sec x \right)\cdot \int{{{\sec }^{2}}x}dx \right\}dx}$  [Apply integration by parts]

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

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

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

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

$=\sec x\tan x+\log \left| \sec x+\tan x \right|-I$              (since, $I=\int{{{\sec }^{3}}xdx}$)

$\Rightarrow 2I=\sec x\tan x+\log \left| \sec x+\tan x \right|+C$

$\Rightarrow I=\frac{1}{2}\left[ \sec x\tan x+\log \left| \sec x+\tan x \right| \right]+C$

Hence, $\int{{{\sec }^{3}}xdx}=\frac{1}{2}\left[ \sec x\tan x+\log \left| \sec x+\tan x \right| \right]+C$, where $C$ is an integration constant.


66. $\int{{{\mathbf{e}}^{\mathbf{2x}}}\mathbf{cos3xdx}}$

Ans. The given integral can be written as

$I=\int{{{e}^{2x}}\cos 3xdx}$ 

$=\frac{1}{3}{{e}^{2x}}\sin 3x-\frac{2}{3}\int{{{e}^{2x}}\sin 3xdx}$          (Applying integration by parts)

$=\frac{1}{3}{{e}^{2x}}\sin 3x-\frac{2}{3}\left( -\frac{1}{3}{{e}^{2x}}\cos 3x+\frac{2}{3}\int{{{e}^{2x}}\cos 3xdx} \right)$ [again, apply integration by parts]

$\Rightarrow I=\frac{1}{3}{{e}^{2x}}\sin 3x+\frac{2}{9}{{e}^{2x}}\cos 3x-\frac{4}{9}\int{{{e}^{2x}}\cos 3xdx}$

$\Rightarrow I=\frac{1}{3}{{e}^{2x}}\sin 3x+\frac{2}{9}{{e}^{2x}}\cos 3x-\frac{4}{9}I$

$\Rightarrow \frac{13}{9}I=\frac{1}{3}{{e}^{2x}}\sin 3x+\frac{2}{9}{{e}^{2x}}\cos 3x$

$\Rightarrow I=\frac{9}{13}\left( \frac{1}{3}{{e}^{2x}}\sin 3x+\frac{2}{9}{{e}^{2x}}\cos 3x \right)+C$

$\Rightarrow \int{{{e}^{2x}}\cos 3xdx}=\frac{{{e}^{2x}}}{13}\left( 3\sin 3x+2\cos 3x \right)+C$, where $C$ is an integration constant.


Integration as an Inverse Differentiation Method

The inverse differentiation process is integration. Instead of distinguishing a function, we are given a function's derivative and asked to find its primitive function, i.e., the original function. This approach is known as integration or anti-differentiation. In fact, there are infinitely many anti-derivatives of each of these functions that can be obtained by arbitrarily choosing C from the real number set. For this purpose, C is typically called an arbitrary constant. In fact, by varying which one gets different antiderivatives (or integrals) of the given function, C is the parameter. 

 

Example:

  1. $\int{(ax^2 + bx + c) dx}$  

Solution: 

$\int{(ax^2 + bx + c)}dx =  \int{ax^2} dx + \int{bx} dx + \int{c} dx$

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

$                                     = a\dfrac{x^2}{3} + b\dfrac{x^2}{2} + cx + C$

So, $\int{(ax^2 + bx + c)}dx = a\dfrac{x^2}{3} + b\dfrac{x^2}{2} + cx + C$

 

Methods of Integration

In the previous paragraph, integrals of certain functions that were readily obtainable from derivatives of certain functions were discussed. It was based on an inspection, i.e. the search for a function F, the derivative of which was f, which led us to the integral of f. However, for many functions, this form, which is based on inspection, is not very suitable. Therefore, by reducing them into standard types, we need to build extra techniques or methods for locating the integrals. Methods that are focused on are popular among them:

1. Integration by substitution

2. Using Partial Fraction Integration

3. Integration by Parts

 

Example: 

  1. Compute the Integral: $\int{e^{\frac{x}{2}}}dx$

Solution: 

Let us equate, $u = \dfrac{x}{2}$

Differentiating both sides,

$\Rightarrow du = \dfrac{dx}{2}$

$\Rightarrow dx = 2du$

On integrating both sides,

$\Rightarrow \int{e^{\frac{x}{2}}}dx = \int{e^u \cdot 2 du}$

$\Rightarrow \int{e^{\frac{x}{2}}}dx = 2 \int{e^u du}$

$\Rightarrow \int{e^{\frac{x}{2}}}dx = 2{e^u} + C$

On putting the value of $u = \dfrac{x}{2}$, we get,

Answer: $\int{e^{\frac{x}{2}}}dx = 2e^{\frac{x}{2}} + C$

 

  1. Calculate the Integral: $\int{\dfrac{dx}{\sqrt{a^2 - x^2}}}$

Solution:

Let us substitute, $u = \dfrac{x}{a}$

Therefore, $x=au$

Differentiating both sides, we get,

$\Rightarrow dx = adu$

Let us use these values in the above equation,

$\Rightarrow \int{\dfrac{dx}{\sqrt{a^2 - x^2}}} = \int{\dfrac{adu}{\sqrt{a^2 - (au)^2}}}$

Let us take $a^2$ common from the denominator, therefore, we get,

$\Rightarrow \int{\dfrac{dx}{\sqrt{a^2 - x^2}}} = \int{\dfrac{adu}{\sqrt{a^2 (1 - (u)^2)}}}$

$\Rightarrow \int{\dfrac{dx}{\sqrt{a^2 - x^2}}} = \int{\dfrac{adu}{a\sqrt{1 - (u)^2}}}$

On cancelling out “a” from numerator and denominator, we get,

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

$\Rightarrow \int{\dfrac{dx}{\sqrt{a^2 - x^2}}} = arc \sin u + C$

Putting the value of $u=\dfrac{x}{a}$, we get,

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

 

  1. Compute $\int{\dfrac{1}{x^2 - 4}} dx$

Solution: 

We will start by factoring the denominator: $(x^2 - 4) = (x + 2)(x - 2)$ 

We will express the fraction as the sum or difference  of the above factors, therefore, we can write it as,

$\dfrac{A}{x+2} + \dfrac{B}{x-2} = \dfrac{1}{x^2 - 4}$

The next step is to simplify the left-hand side by taking the LCM of $(x+2)$ and $(x-2)$

$\dfrac{A}{x+2} + \dfrac{B}{x-2} = \dfrac{A(x-2)+B(x+2)}{x^2-4}$

On equating the above equation to the right-hand side, we get

$A(x-2)+B(x+2)=1$

The above equation is about two functions; for all values of $x$, the two sides have to be equal. Let's substitute $x= 2$ for both of them.

$A(0)+B(4)=1$

$B=\dfrac{1}{4}$

Similarly, on substituting $x=-2$, we get,

$A(-4)+B(0)=1$

$A(-4)=1$

$A=-\dfrac{1}{4}$

Therefore, on putting the value of $A$ and $B$ in $\dfrac{A}{x+2} + \dfrac{B}{x-2} = \dfrac{1}{x^2-4}$, we get

$\int{\dfrac{1}{x^2 - 4}}dx = \int{\dfrac{\dfrac{-1}{4}}{x+2} + \dfrac{\dfrac{1}{4}}{x-2}}dx$

$\int{\dfrac{1}{x^2 - 4}}dx = -\dfrac{1}{4} \int{\dfrac{1}{x+2}}dx + \dfrac{1}{4}\int{\dfrac{1}{x-2}}dx$

$\int{\dfrac{1}{x^2 - 4}}dx = -\dfrac{1}{4} \int{\dfrac{1}{x+2}}dx + \dfrac{1}{4}\int{\dfrac{1}{x-2}}dx$

Answer $= \int{\dfrac{1}{x^2 - 4}}dx = -\dfrac{1}{4} ln|x+2| + \dfrac{1}{4} ln|x-2|+C$

 

  1. What is $\int{\dfrac{ln{x}}{x^2}}dx$?

Assign the value of $u = ln(x)$

On differentiating, $\dfrac{du}{dx} = \dfrac{1}{x}$

  $v = -\dfrac{1}{x}$

On differentiating, $dv= \dfrac{1}{x^2}dx$

Putting these values in the given equation,

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

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

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

 

Integrals of Some Particular Functions

A rational function is defined in the form $\dfrac{P(x)}{Q(x)}$ as the ratio of two polynomials, where $P(x)$ and $Q(x)$ are polynomials of $x$ and $Q(x)$ of 0.0. If the $P(x)$ degree is lower than the $Q(x)$ degree, then the rational function is called valid, otherwise, it is called improper.

Integral Function

Integral Value

$\int{\dfrac{dx}{x^2 - a^2}}$

$\dfrac{1}{2a} log \left|\dfrac{x-a}{x+a}\right| + C$

$\int{\dfrac{dx}{a^2 - x^2}}$

$\dfrac{1}{2a} log \left|\dfrac{a+x}{a-x}\right| + C$

$\int{\dfrac{dx}{x^2 + a^2}}$

$\dfrac{1}{a} \tan^{-1}\left(\dfrac{x}{a}\right) + C$

$\int{\dfrac{dx}{x^2 - a^2}}$

$log\left|x+\sqrt{x^2 - a^2}\right| + C$

$\int{\dfrac{dx}{a^2 - x^2}}$

$\sin^{-1}\left(\dfrac{x}{a}\right)+C$

$\int{\dfrac{dx}{x^2 + a^2}}$

$log\left|x+\sqrt{x^2 + a^2}\right| + C$

 

Integration by Partial Fractions

If the integrand (the expression after the integral sign) is in the shape of an algebraic fraction and the integral can not be evaluated by simple methods, before integration takes place, the fraction must be expressed in partial fractions. The steps necessary to decompose an algebraic fraction into its partial fractions result from the reverse process being taken into account (or subtraction). We decompose fractions into certain partial fractions because:

  • it makes it much simpler to do such integrals, and

  • in the Laplace transformation, which we meet later, it is used.

The numerator must be at least one degree lower than the denominator before a fractional function can directly be represented in partial fractions.

There is a partial fraction of the form $\dfrac{A}{(ax+b)}$ for each linear factor $(ax+b)$ in the denominator of a rational fraction, where A is a constant. If a linear factor is repeated n times in the denominator, n corresponding partial fractions from degree 1 to n will be present.

 

Integration by Parts

If u and v are two distinguishable functions of a single x variable (say). Then we have, by the product law of differentiation, 

$\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

The product integral of two functions = (first function) $\times$ (integral function of Second function) $-$ Integral of ((first function differential coefficient) $\times$ (second function integral))

 

Definite Integral

The definite integral has a distinctive meaning. A definite integral is denoted by $\int_{a}^{b}f(x) dx$ , where an is considered

the integral's lower limit, and b is called the integral's upper limit. The definite integral is either added as the limit of a sum or if in the interval $[a, b]$ it has an antiderivative $F$, then its value is the difference between F values at the endpoints, i.e. $F(b)-F (a)$.

 

Example:

  1. $\int_{-1}^{1}\sin^5x \cdot \cos^4x dx$

Solution

Let, $f(x) = \sin^5x \cdot \cos^4x dx$

$f(-x)=\sin^{5}(-x) \cdot \cos^{-4}(-x) dx$

$           = -\sin^5x \cdot \cos^4x$

$           = -f(x) $

From the above, it can be deduced that $f(x)$ is an odd function.

Therefore, $f(x)=\int_{-1}^{1}\sin^5x \cdot \cos^4x dx=0$

 

  1. $\int_{-4}^{4}\sin^2x dx$

Solution:

Let, $f(x) = \sin^2x dx$

$f(-x)=\sin^2(-x)$

$           =\sin^2x$

$           =f(x)$

Therefore, $f(x)$ is an even function.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x dx = 2\int_{0}^{\frac{\pi}{4}}\sin^2x dx$

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x dx = 2\int_{0}^{\frac{\pi}{4}}\left(\dfrac{1-\cos2x}{2}\right)dx$

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x dx = \left[x - \dfrac{\sin2x}{2}\right]_{0}^{\frac{\pi}{4}}$

Answer $= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x dx = \left[\dfrac{\pi}{4} - \dfrac{1}{2}\right]$

 

Conclusion

The Differential Calculus focuses on the derivative notion. The problem of defining tangent lines for the function graphs and calculating the slope of such lines was the original motivation for the derivative. The problem of defining and calculating the area of the region bounded by the graph of the functions motivates Integral Calculus. As we all know, one of the essential phases in career-making is Class 12 Maths Chapter 7 Extra Questions, and the marks earned for Class 12th are also responsible for college admission. We have therefore concentrated on preparing you by providing you with various sets of questions that can be asked in your twelfth board. You will get Class 12 Maths Integration Important Questions here at Vedantu so that you can download the files and start practising.

Integration Important Questions Class 12 covers important concepts such as integrations, definite and indefinite integrals, certain definite integral properties, basic calculus theorem, and also integration methods such as:

  • Integration by parts

  • Integration by substitution

  • Integration using partial fractions

Here, integrals are given. Students should be comprehensive with all the definitions when studying for the Class 12 Maths board examination. If the principles are straightforward, they should be able to solve any problem. Students need thorough preparation for that. Here, as per the CBSE Board syllabus, the relevant questions are taken up.


What are the Benefits of Important Questions from Vedantu for Class 12 Chapter 7 - Integrals

Unravel the key to conquering the challenges posed by Class 12 Integrals through Vedantu’s curated collection of Important Questions. Delve into a wealth of beneficial queries designed to deepen your understanding of this crucial chapter. Explore the advantages of these specifically chosen questions, tailored to bolster your grasp on Integrals, paving the way for a confident and successful mastery of Mathematics.


  • Focus on key topics for efficient studying.

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  • Teaches effective time management.

  • Enables self-assessment and progress tracking.

  • Strategic approach for higher scores.

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Conclusion

Differential Calculus delves into the derivative concept, initially motivated by defining tangent lines and calculating their slopes. Integral Calculus, on the other hand, emerges from the need to define and compute the area bounded by function graphs. Recognizing the crucial role of Class 12 Maths Chapter 7, Extra Questions, in shaping careers and influencing college admissions, we've focused on your preparation. Vedantu provides diverse sets of questions for your twelfth board, ensuring you're well-equipped. Access Class 12 Maths Integration Important Questions here, enabling you to download and commence practice for a successful academic journey.


Important Related Links for CBSE Class 12 Maths 

FAQs on Important Questions for CBSE Class 12 Maths Chapter 7- Integrals 2023-24

1. Which topics are important in Chapter 7 of Class 12 Maths ?

Students need to have an idea of which the most important chapters are and their respective weightage for the exam and then study accordingly. 

  • Definite and Indefinite Integrals

  • Integration reverse process of differentiation

  • Theorem of Calculus

  • Methods of Integration

  • Properties of Integrals

  • The connection between definite and indefinite integrals

One useful material that they could use to determine this is the important questions which have all the important topics covered. That being said, Calculus is the most important chapter.


2. Is Chapter 7 of Class 12 Maths difficult?

Truth to be told, the difficulty of Chapter 7 of Class 12 Maths depends upon the amount of preparation that the students have done for the exam. The chapter on Integrals is a new concept to students, and without the right focus on the important areas, there is no chance of students scoring good marks. The important questions prepared are an essential guide that students can use to learn areas from this chapter that have the most weightage.

3. How can I get full marks for questions from Chapter 7 of Class 12 Maths?

Time, patience, practice, dedication, and hard work with help from the important questions provided by Vedantu will do the trick in helping students achieve their perfect score of 100. Students need to remember that cracking this chapter depends on their ability to solve the most important concepts. Therefore, students can use these questions to prepare and solve the most difficult questions as these are specially prepared by experts following the question paper difficulty. Thus, be on the way to full marks!

4. How can I study Chapter 7 of Class 12 Maths?

Chapter 7 of Class 12 Maths is all about Integration and for different students, the difficulty level of the chapter is different. The following points highlight how all students alike can prepare this chapter: 

  • Following the exam pattern

  • Preparing notes

  • Practicing and solving problems

  • Creating a study schedule for the chapter

  • Using NCERT Solutions, Revision notes, Important Questions and other references

  • Attempting and solving previous years’ question papers

  • Clearing every existing doubt before the exams

5. What are the benefits of using important questions for Chapter 7 of Class 12 Maths?

Integration is one of the most important chapters that students have to learn and then face questions in their Class 12. So, they really need to focus on whatever is important in the chapter. Using the Important Questions provided by Vedantu(vedantu.com) at free of cost, students can learn about the concepts and properties of integrals in more detail as they are emphasized because of their importance. These are available on mobile app also. At the same time, every concept is provided in an easy and simple way through professionally drafted answers.