Integrals Class 12 Notes CBSE Maths Chapter 7 (Free PDF Download)

1. Find $\int(1-x) \sqrt{x} d x$.

Ans. $\int(1-x) \sqrt{x} d x=\int(\sqrt{x}-x \sqrt{x}) d x$

$=\int\left(x^{1 / 2}-x^{3 / 2}\right)d x=\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}+C \\$ 

$\left [ \because \int x^n d x=\frac{x^{n+1}}{n+1},n\neq -1 \right ]$

2. Evaluate $\int \frac{2}{1+\cos 2 x} d x$.

Ans. $\int \frac{2}{1+\cos 2 x} d x=\int \frac{2}{2 \cos ^2 x} d x$

$$=\int \sec ^2 x d x=\tan x+C .$$

3. Find $\int_4^5 e^x d x$.

Ans. $\int_4^5 e^x d x=\left[e^x\right]_4^5=e^5-e^4=e^4(e-1)$.

4. Find $\int_0^{\pi / 4} \tan x d x$.

Ans. $\int_0^{\pi / 4} \tan x d x=[\log |\sec x|]_0^{\pi / 4}$

$$ \begin{aligned} & =\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0| \\ & =\log \sqrt{2}-\log 1 \\ & =\log \sqrt{2} . \end{aligned} $$

5. Evaluate $\int_0^1 \frac{d x}{\sqrt{1-x^2}}$

Ans. $\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1$

$$\begin{aligned}& =\sin ^{-1}(1)-\sin ^{-1}(0) \\& =\frac{\pi}{2}-0=\frac{\pi}{2}\end{aligned}$$

Section-B (2 Marks Questions)

6. Evaluate $\int\left(\operatorname{cosec}^2 x-\cot x\right) e^x d x$.

Ans.  $\int\left(\operatorname{cosec}^2 x-\cot x\right) e^x d x$

$=-\int\left(\cot x-\operatorname{cosec}^2 x\right) e^x d x \\$

$=-\int\left[\cot x+\left(-\operatorname{cosec}^2 x\right)\right] e^x d x=-e^x \cot x+C \\$

${\left[\therefore  \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x \cdot f(x)\right]}$

7. Evaluate $\int \sin ^3 x d x$.

Ans.\begin{aligned} & \int \sin ^3 x d x=\int \frac{3 \sin x-\sin 3 x}{4} d x \\ &= \frac{1}{4} \int 3 \sin x d x-\frac{1}{4} \int \sin 3 x d x \\ &= \frac{1}{4}\left(-3 \cos x+\frac{\cos 3 x}{3}\right)+C . \end{aligned}

8. Evaluate $\int \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x$.

Ans. $\int \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x=\int \sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} d x$

$$ \begin{aligned} & =\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0| \\ & =\log \sqrt{2}-\log 1 \\ & =\log \sqrt{2} . \end{aligned} $$

9. Write the value of $\int \frac{2-3 \sin x}{\cos ^2 x} d x$.

Ans.

$\int \frac{2-3 \sin x}{\cos ^2 x} d x$

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

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

$=2 \int \sec ^2 x d x-3 \int \sec x \tan x d x \\$ 

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

10. Evaluate $\int \frac{2 \cos x}{\sin ^2 x} d x$

Ans. $\int \frac{2 \cos x}{\sin ^2 x} d x=\int 2 \operatorname{cosec} x \cot x d x$

$=-2 \operatorname{cosec} x+C \\

$\left [ \because \int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x \right ]$

11. Write the value of $\int_{\frac{-\pi}{4}}^0 \frac{1+\tan x}{1-\tan x} d x$

$$I=\int_{\frac{-\pi}{4}}^0 \frac{1+\tan x}{1-\tan x} d x=\int_{\frac{-\pi}{4}}^0 \tan\left(\frac{\pi}{4}+x\right) d x$$

Ans. $\begin{aligned}& =\left[\log \left|\sec\left(\frac{\pi}{4}+x\right)\right|\right]_{\dfrac{-\pi}{4}}^0 \\& =\log \sqrt{2}=\frac{1}{2} \log 2\end{aligned}$

12. Find $\int \log x d x$.

Ans. $\int(\log x .1) d x=\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x=(\log x) x-\int \frac{1}{x} x d x=x \log x-x+C$.

13. If $\int_0^a \frac{1}{4+x^2} d x=\frac{\pi}{8}$, then find the value of $a$.

Ans.

We have, $\int_0^a \frac{1}{4+x^2} d x=\frac{\pi}{8}$

$$\begin{aligned} & \Rightarrow \frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_0^a=\frac{\pi}{8} \\ & \Rightarrow \tan ^{-1} \frac{a}{2}-0=\frac{\pi}{4} \\ & \Rightarrow \frac{a}{2}=1 \Rightarrow a=2 \\ & \Rightarrow \tan ^{-1} \frac{a}{2}=\frac{\pi}{4} \end{aligned}$$

PDF Summary - Class 12 Maths Integrals Notes (Chapter 7)

Differentiation is the inverse of integration. Integration is the process of determining a function whose differential coefficient is known.

So from the above, if the differential coefficient of ${\text{F}}({\text{x}})$ is $f({\text{x}})$

i.e. $\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] = f({\text{x}})$, then one can say that the antiderivative or integral of $f(x)$ is $F(x)$, written as $\int f (x)dx = F(x)$,

Here $\int {\text{d}} {\text{x}}$ is the notation of integration $f({\text{x}})$ is the integrand, $x$ is the variable of integration and dx denotes the integration with respect to ${\text{X}}$.

Indefinite Integral:

We know that if $\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] = f({\text{x}})$, then $\int f ({\text{x}}){\text{dx}} = {\text{F}}({\text{x}})$.

Also, for any arbitrary constant C,

$\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}}) + {\text{C}}] = \dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] + 0 = f({\text{x}})$

$\therefore \int f ({\text{x}}){\text{dx}} = {\text{F}}({\text{x}}) + {\text{C}}$

This shows that ${\text{F}}({\text{x}})$ and ${\text{F}}({\text{x}}) + {\text{C}}$ are both integrals of the same function ${\text{f}}({\text{x}})$. Thus, for different values of ${\text{C}}$, we obtain different integrals of ${\text{f}}({\text{x}})$. This implies that the integral of ${\text{f}}({\text{x}})$ is not definite. By virtue of this property ${\text{F}}({\text{x}})$ is called the indefinite integral of ${\text{f}}({\text{x}})$.

1.1 Properties of Indefinite Integration

1. $\quad \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\int {\text{f}} ({\text{x}}){\text{dx}}} \right] = f({\text{x}})$

2. $\int {{f^\prime }} (x){\text{dx}} = \int {\dfrac{{\text{d}}}{{{\text{dx}}}}} [f({\text{x}})]{\text{dx}} = f({\text{x}}) + {\text{c}}$

3. $\quad \int {\text{k}} f({\text{x}}){\text{dx}} = {\text{k}}\int f ({\text{x}}){\text{dx}}$, where ${\text{k}}$ is any constant

4. If ${f_1}({\text{x}}),{f_2}({\text{x}}),{f_3}({\text{x}}), \ldots $ (finite in number) are functions of ${\text{x}}$, then

$ \int{\left[ {{f}_{1}}(x)\pm {{f}_{2}}(x)\pm {{f}_{3}}(x)\ldots  \right]}dx $

$   =\int{{{f}_{1}}}(x)dx\pm \int{{{f}_{2}}}(x)dx\pm \int{{{f}_{3}}}(x)dx\pm \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ } $

5. If $\int f (x)dx = F(x) + c$ then $\int f (ax \pm b)dx = \dfrac{1}{a}F(ax \pm b) + c$

1.2 Standard Formula of Integration

The definition of an integral has the following conclusions as a direct result.

1. $\quad \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne  - 1$.

2. $\quad \int {\dfrac{1}{x}} dx = \log |x| + C$

3. $\int {{{\text{e}}^{\text{x}}}} {\text{dx}} = {{\text{e}}^{\text{x}}} + {\text{C}}$

4. $\int {{{\text{a}}^{\text{x}}}} {\text{dx}} = \dfrac{{{{\text{a}}^{\text{x}}}}}{{{{\log }_{\text{e}}}{\text{a}}}} + {\text{C}}$.

5. $\quad \int {\sin } xdx =  - \cos x + C$

6. $\quad \int {\cos } xdx = \sin x + C$

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

8. $\quad \int {{{\operatorname{cosec} }^2}} xdx =  - \cot x + C$

9. $\quad \int {\sec } x\tan xdx = \sec x + C$

10. $\int {\operatorname{cosec} } x\cot xdx =  - \operatorname{cosec} x + C$.

11. $\int {\tan } xdx =  - \log |\cos x| + C = \log |\sec x| + C$.

12. $\int {\cot } xdx = \log |\sin x| + C$

13. $\int {\sec } xdx = \log |\sec x + \tan x| + C$

14. $\int {\operatorname{cosec} } xdx = \log |\operatorname{cosec} x - \cot x| + C$

15. $\int {\dfrac{{{\text{dx}}}}{{\sqrt {1 - {{\text{x}}^2}} }}}  = {\sin ^{ - 1}}{\text{x}} + {\text{C}};|{\text{x}}| < 1$

16. $\int {\dfrac{{dx}}{{1 + {x^2}}}}  = {\tan ^{ - 1}}x + C$

17. $\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\sqrt {{{\text{x}}^2} - 1} }}}  = {\sec ^{ - 1}}|{\text{x}}| + {\text{C}};|{\text{x}}| > 1$

2. Methods of Integration

2.1 Method of Substitution

By using the suitable substitution, the variable $x$ in $\int f (x)dx$ is changed into another variable $t$ so that the integrand $f(x)$ is changed into ${\text{F}}({\text{t}})$ is an algebraic sum of standard integrals or a standard integral. There is no universal formula for determining a suitable substitute, and experience is the greatest guidance in this regard.

The Following Ideas, on the Other Hand, will be Beneficial.

1. If the integrand is of the form ${f^\prime }(ax + b)$, then we

put $ax + b = t$ and $dx = \dfrac{1}{a}dt$

Thus, $\int {{f^\prime }} (ax + b)dx = \int {{f^\prime }} (t)\dfrac{{dt}}{a}$

$ = \dfrac{1}{a}\int {{f^\prime }} (t)dt = \dfrac{{f(t)}}{a} = \dfrac{{f(ax + b)}}{a} + c$

2. When the integrand is of the form ${x^{n - 1}}{f^\prime }\left( {{x^n}} \right)$, we put ${x^n} = t$ and $n{x^{n - 1}}dx = dt$

Thus, $\int {{x^{n - 1}}} {f^\prime }\left( {{x^n}} \right)dx = \int {{f^\prime }} (t)\dfrac{{dt}}{n} = \dfrac{1}{n}$

$\int {{f^\prime }} ({\text{t}}){\text{dt}} = \dfrac{1}{{\text{n}}}f({\text{t}}) = \dfrac{1}{{\text{n}}}f\left( {{{\text{x}}^{\text{n}}}} \right) + {\text{c}}$

3. When the integrand is of the form ${[f(x)]^n} \cdot {f^\prime }(x)$, we put $f(x) = t$ and ${f^\prime }(x)dx = dt$

Thus, ${\text{(Tex translation failed)}}$

4. When the integrand is of the form $\dfrac{{{f^\prime }(x)}}{{f(x)}}$, we put

$f({\text{x}}) = {\text{t}}$ and ${f^\prime }({\text{x}}){\text{dx}} = {\text{dt}}$

Thus, $\int {\dfrac{{{f^\prime }({\text{x}})}}{{f({\text{x}})}}} {\text{dx}} = \int {\dfrac{{{\text{dt}}}}{{\text{t}}}}  = \log |{\text{t}}| = \log |f({\text{x}})| + {\text{c}}$

2.1.1 Some Special Integrals

1. $\int {\dfrac{{dx}}{{{x^2} + {a^2}}}}  = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$

2. $\quad \int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^2} - {{\text{a}}^2}}}}  = \dfrac{1}{{2{\text{a}}}}\log \left| {\dfrac{{{\text{x}} - {\text{a}}}}{{{\text{x}} + {\text{a}}}}} \right| + {\text{C}}$

3. $\quad \int {\dfrac{{dx}}{{{a^2} - {x^2}}}}  = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + C$

4. $\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }}}  = {\sin ^{ - 1}}\dfrac{x}{a} + C$

5. $\quad \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }}}  = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$

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

7. $\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$.

8. $\quad \int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}}  + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$

9. $\quad \int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$

2.1.2 Integrals of the Form

1. $\int f \left( {{{\text{a}}^2} - {{\text{x}}^2}} \right){\text{dx}}$,

2. $\int f \left( {{a^2} + {x^2}} \right)dx$,

3. $\int f \left( {{x^2} - {a^2}} \right)dx$,

4. $\int f \left( {\dfrac{{a - x}}{{a + x}}} \right)dx$,

Working Rule

Integral Substitution

$\int f \left( {{{\text{a}}^2} - {{\text{x}}^2}} \right){\text{dx}},\quad {\text{x}} = {\text{a}}\sin \theta $ or ${\text{x}} = {\text{a}}\cos \theta $

$\int f \left( {{a^2} + {x^2}} \right)dx,\quad x = a\tan \theta $ or $x = a\cot \theta $

$\int f \left( {{x^2} - {a^2}} \right)dx,\quad x = a\sec \theta $ or $x = a\operatorname{cosec} \theta $

$\int f \left( {\dfrac{{a - x}}{{a + x}}} \right)dx$ or $\int f \left( {\dfrac{{a + x}}{{a - x}}} \right)dx\quad x = a\cos 2\theta $

2.1.3 Integrals of the Form

1. $\int {\dfrac{{dx}}{{a{x^2} + bx + c}}} $

2. $\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} $

3. $\int {\sqrt {a{x^2} + bx + c} } dxa$,

Working Rule

(i)Make the coefficient of ${x^2}$ unity by taking the coefficient of ${{\text{x}}^2}$ outside the quadratic.

(ii) Complete the square in the terms involving $x$, i.e. write $a{x^2} + bx + c$ in the form

$a\left[ {{{\left( {x + \dfrac{b}{{2a}}} \right)}^2}} \right] - \dfrac{{\left( {{b^2} - 4ac} \right)}}{{4a}}$

(iii) One of the nine special integrals is used to transform the integrand.

(iv) Then just integrate the function.

2.1.4 Integrals of the Form

1. $\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$

2. $\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} {\text{dx}}$

3. $\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$

Integral Working Rule

$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} {\text{dx}}$ Put ${\text{px}} + {\text{q}} = \lambda (2{\text{ax}} + {\text{b}}) + \mu $ or ${\text{px}} + {\text{q}} = \lambda $ (derivative of quadratic) $ + \mu $.

When the coefficient of x and the constant term on both sides are compared, we obtain

${\text{p}} = 2{\text{a}}\lambda $ and ${\text{q}} = {\text{b}}\lambda  + \mu  \Rightarrow \lambda  = \dfrac{{}}{{2{\text{a}}}}$ and $\mu  = \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)$. 

Then integral becomes

$\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$

$ = \dfrac{{2{\text{ax}} + {\text{b}}}}{{2{\text{a}}}}\int {\dfrac{{{\text{dx}}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} {\text{dx}} + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\dfrac{{\text{d}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} $

$ = \dfrac{{dx}}{{2a}}\log \left| {a{x^2} + bx + c} \right| + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{a{x^2} + bx + c}}} $

$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} {\text{dx}}$ 

In this case the integral becomes

$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} dx = $

$\dfrac{p}{{2a}}\int {\dfrac{{2ax + b}}{{\sqrt {a{x^2} + bx + c} }}} dx + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $

$ = \dfrac{p}{a}\sqrt {a{x^2} + bx + c}  + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $

$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$

In this example, the integral is transformed to

$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx = \dfrac{p}{{2a}}\int {(2ax + b)} \sqrt {a{x^2} + bx + c} dx$

$ + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} } {\text{dx}}$

$ = \dfrac{{\text{p}}}{{3{\text{a}}}}{\left( {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} \right)^{3/2}} + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} } {\text{dx}}$

2.1.5 Integrals of the Form

$\int {\dfrac{{{\text{P}}({\text{x}})}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} {\text{dx}}$, where ${\text{P}}({\text{x}})$ is a polynomial in ${\text{x}}$ of degree ${\text{n}} \geqslant 2$.

Working Rule: Write

$  {\int {\dfrac{{{\text{P}}({\text{x}})}}{{\sqrt {a{x^2} + bx + c} }}} dx = } $

$  { = \left( {{a_0} + {a_1}x + {a_2}{x^2} +  \ldots  + {a_{n - 1}}{x^{n - 1}}} \right)\sqrt {a{x^2} + bx + c}  + k\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} } $

where ${\text{k}},{{\text{a}}_0},{{\text{a}}_1}, \ldots {{\text{a}}_{{\text{n}} - 1}}$ are constants that may be found by separating the above relation and equating the coefficients of various powers of$x$ on both sides.

2.1.6 Integrals of the Form

$\int {\dfrac{{{x^2} + 1}}{{{x^4} + k{x^2} + 1}}} dx\quad $ or $\quad \int {\dfrac{{{x^2} - 1}}{{{x^4} + k{x^2} + 1}}} dx$

where ${\text{k}}$ is a constant positive, negative or zero.

Working Rule

(i) ${{\text{x}}^2}$ is divided into the numerator and denominator.

(ii) Put $x - \dfrac{1}{x} = z$ or $x + \dfrac{1}{x} = z$, Whatever substitution yields the numerator of the resultant integrand on differentiation

(iii) In ${\text{Z}}$, evaluate the resultant integral.

(iv) In terms of X, express the outcome.

2.1.7 Integrals of the Form

$\int {\dfrac{{{\text{dx}}}}{{{\text{P}}\sqrt {\text{Q}} }}} $, where ${\text{P}},{\text{Q}}$ are linear or quadratic functions of ${\text{x}}$.

Integral Substitution

$\int {\dfrac{1}{{(ax + b)\sqrt {cx + d} }}} dx$

$cx + d = {z^2}$

$\int {\dfrac{{dx}}{{\left( {a{x^2} + bx + c} \right)\sqrt {px + q} }}} \quad px + q = {z^2}$

$\int {\dfrac{{dx}}{{(px + q)\sqrt {a{x^2} + bx + c} }}} \quad px + q = \dfrac{1}{z}$

$\int {\dfrac{{dx}}{{\left( {a{x^2} + b} \right)\sqrt {c{x^2} + d} }}} \quad x = \dfrac{1}{z}$

3. Method of Partial Fractions For Rational Functions

By resolving the integrand into partial fractions, Integrals of the type $\int {\dfrac{{p(x)}}{{g(x)}}} $  may be integrated. 

The Following is How we Proceed:

First of all check the degree of $p({\text{x}})$ and $g({\text{x}})$.

If degree of $p({\text{x}}) \geqslant $ degree of $g({\text{x}})$, then divide $p({\text{x}})$ by $g$(x) till its degree is less, i.e. put in the form $\dfrac{{p({\text{x}})}}{{g({\text{x}})}} = {\text{r}}({\text{x}}) + \dfrac{{f({\text{x}})}}{{g({\text{x}})}}$ where degree of $f({\text{x}}) < $ degree of$g({\text{x}})$

CASE 1: When there are non-repeated linear components in the denominator. i.e.

$g({\text{x}}) = \left( {{\text{x}} - {\alpha _1}} \right)\left( {{\text{x}} - {\alpha _2}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$

In such a case write $f({\text{x}})$ and $g$ as:

$\dfrac{{f({\text{x}})}}{{g({\text{x}})}} = \dfrac{{{{\text{A}}_1}}}{{\left( {{\text{x}} - {\alpha _1}} \right)}} + \dfrac{{{{\text{A}}_2}}}{{\left( {{\text{x}} - {\alpha _2}} \right)}} +  \ldots  + \dfrac{{{{\text{A}}_{\text{n}}}}}{{\left( {{\text{x}} - {\alpha _{\text{n}}}} \right)}}$

After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.

CASE 2: When there are both repeated and non-repeated linear factors in the denominator.

i.e.$g({\text{x}}) = {\left( {{\text{x}} - {\alpha _1}} \right)^2}\left( {{\text{x}} - {\alpha _3}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$

In such a case write $f(x)$ and $g(x)$ as:

$\dfrac{{f({\text{x}})}}{{g({\text{x}})}} = \dfrac{{{{\text{A}}_1}}}{{{\text{x}} - {\alpha _1}}} + \dfrac{{{{\text{A}}_2}}}{{{{\left( {{\text{x}} - {\alpha _1}} \right)}^2}}} + \dfrac{{{{\text{A}}_3}}}{{{\text{x}} - {\alpha _3}}} +  \ldots  + \dfrac{{{{\text{A}}_{\text{n}}}}}{{\left( {{\text{x}} - {\alpha _{\text{n}}}} \right)}}$

After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.

Note: corresponding to a linear factor that has been repeated${({\text{x}} - {\text{a}})^{\text{r}}}$ in the denominator, a sum of $r$ partial fractions of the type $\dfrac{{{{\text{A}}_1}}}{{{\text{x}} - {\text{a}}}} + \dfrac{{{{\text{A}}_2}}}{{{{({\text{x}} - {\text{a}})}^2}}} +  \ldots  + \dfrac{{{{\text{A}}_{\text{r}}}}}{{{{({\text{x}} - {\text{a}})}^r}}}$ is taken.

CASE 3: When there is a non-repeated quadratic component in the denominator that cannot be factored further:

$g({\text{x}}) = \left( {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} \right)\left( {{\text{x}} - {\alpha _3}} \right)\left( {{\text{x}} - {\alpha _4}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$

In such a case express $f(x)$ and $g(x)$ as:

$\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}x + {A_2}}}{{a{x^2} + bx + c}} + \dfrac{{{A_3}}}{{x - {\alpha _3}}} +  \ldots  + \dfrac{{{A_n}}}{{x - {\alpha _n}}}$

After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.

CASE 4: When there is a repeating quadratic component in the denominator that cannot be factored further:

i.e. $g(x) = {\left( {a{x^2} + bx + c} \right)^2}\left( {x - {\alpha _5}} \right)\left( {x - {\alpha _6}} \right) \ldots \left( {x - {\alpha _n}} \right)$

In such a case write $f({\text{x}})$ and $g({\text{x}})$ as

$  {\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}x + {A_2}}}{{a{x^2} + bx + c}} + \dfrac{{{A_3}x + {A_4}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + } $

$  {\dfrac{{{A_5}}}{{x - {\alpha _5}}} +  \ldots  + \dfrac{{{A_n}}}{{\left( {x - {\alpha _n}} \right)}}} $

After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.

CASE 5: If $x$ will be the even power then

(i) Put ${{\text{x}}^2} = {\text{z}}$ in the integrand.

(ii) Resolve the resulting rational expression in ${\text{z}}$ into partial fractions

(iii) Put ${\text{z}} = {{\text{x}}^2}$ again in the partial fractions and then integrate both sides.

4. Method of Integration by Parts

Integration by parts refers to the process of combining the output of two functions.

e.g. if $u$ and $v$ are two functions of $x$, then $\int {(uv)} {\text{dx}} = u \cdot \int v \;{\text{dx}} - \int {\left( {\dfrac{{{\text{d}}u}}{{{\text{dx}}}} \cdot \int v {\text{dx}}} \right)} {\text{dx}}$

To put it another way, the integral of the product of two functions = first function $ \times $ integral of the second - integral of (differential of first $ \times $ integral of the second function).

Working Hints

1. Choose the first and second functions in such a way that the derivative and integral of the first function may be determined quickly.

2. In case of integrals of the form $\int f ({\text{x}}) \cdot {{\text{x}}^{\text{n}}}{\text{dx}}$, take ${{\text{x}}^{\text{n}}}$ as the first function and $f({\text{x}})$ as the second function.

3. In case of integrals of the form $\int {{{(\log x)}^{\text{n}}}}  \cdot 1{\text{dx}}$, take 1 as the second function and ${(\log x)^n}$ as the first function.

4. The rule of parts integration can be applied several times if necessary.

5. If the two functions are of different types, the first function can be chosen as the one with the first initial in the word "ILATE," where

I - Inverse Trigonometric function

L - Logarithmic function

A - Algebraic function ${\text{T}} - $ Trigonometric function

${\text{E}} - $ Exponential function.

6. If both functions are trigonometric, use the second function as the second function with a simple integral. If both functions are algebraic, choose the one with the simpler derivative as the initial function.

7. If the integral is an inverse trigonometric function of an algebraic expression in x, simplify the integrand first using an appropriate trigonometric substitution before integrating the new integrand.

4.1 Integrals of the Form $\int {{{\text{e}}^x}} \left[ {f({\text{x}}) + {f^\prime }({\text{x}})} \right]{\text{dx}}$

Working Rule

(i) Divide the integral into two equal halves.

(ii) Only use parts to integrate the first integral, i.e.

$ = \left[ {f({\text{x}}) \cdot {{\text{e}}^{\text{x}}} - \int {{f^\prime }} ({\text{x}}) \cdot {{\text{e}}^{\text{x}}}{\text{x}}} \right] + \int {{{\text{e}}^{\text{x}}}} {f^\prime }({\text{x}}){\text{dx}}$

$ = {{\text{e}}^{\text{x}}}f({\text{x}}) + {\text{C}}$

4.2 Integrals of the Form:

After integrating by parts, the initial integrand reappears.

Working Rule

(i) Use the part-by-part integration technique twice.

(ii) If we integrate by parts a second time, we'll get the same integrand as before, therefore we'll set it equal to $I$.

(iii) On one side, transpose and gather words involving $I$, and assess $I$.

5. Integral of the Form (Trigonometric Formats)

5.1 

1. $\int {\dfrac{{dx}}{{a + b\cos x}}} $

2. $\int {\dfrac{{dx}}{{a + b\sin x}}} $

3. $\int {\dfrac{{dx}}{{a + b\cos x + c\sin x}}} $

Working Rule

(i) Put $\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$ and $\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$ so that the given integrand becomes a function of $\tan \dfrac{x}{2}$.

(ii) Put $\tan \dfrac{{\text{x}}}{2} = {\text{z}} \Rightarrow \dfrac{1}{2}{\sec ^2}\dfrac{{\text{x}}}{2}{\text{dx}} = {\text{dz}}$

(iii) Integrate the resulting rational algebraic function of $z$

(iv) In the answer, put $z = \tan \dfrac{x}{2}$.

5.2 Integrals of the Form

1. $\int {\dfrac{{dx}}{{a + b{{\cos }^2}x}}} $

2. $\int {\dfrac{{dx}}{{a + b{{\sin }^2}x}}} $

3. $\int {\dfrac{{dx}}{{a{{\cos }^2}x + b\sin x\cos x + {{\operatorname{csin} }^2}x}}} $

Working Rule

(i) Divide the numerator and denominator by ${\cos ^2}x$.

(ii) In the denominator, replace sec'x, if any, by $1 + {\tan ^2}x$.

(iii) Put $\tan x = z \Rightarrow {\sec ^2}xdx = dz$

(iv) Integrate the resulting rational algebraic function of $z$.

(v) In the answer, put $z = \tan x$. 

5.3 Integrals of the Form

$\int {\dfrac{{a\cos x + b\sin x}}{{c\cos x + d\sin x}}} dx$

Working Rule

(i) Put Numerator $ = \lambda $ (denominator) $ + \mu $ (derivative of denominator) $a\cos x + b\sin x = \lambda (c\cos x + d\sin x) + \mu ( - c\sin x + d\cos x)$

(ii) Equate coefficients of $\sin x$ and $\cos x$ on both sides and find the values of $\lambda $, and $\mu $.

(iii) Divide the provided integral into two parts and evaluate each independently, as follows:

$\int {\dfrac{{a\cos x + b\sin x}}{{c\cos x + d\sin x}}} dx = $

$\lambda \iint {{\text{ld}}}{\text{x}} + \mu \int {\dfrac{{ - {\text{csinx}} + {\text{d}}\cos {\text{x}}}}{{{\text{a}}\cos {\text{x}} + {\text{b}}\sin {\text{x}}}}} {\text{dx}} = \lambda {\text{x}} + \mu \log |{\text{a}}\cos {\text{x}} + {\text{b}}\sin {\text{x}}|$

(iv) Substitute the values of $\lambda $ and $\mu $ found in step 2 . 

5.4 Integrals of the Form $\int {\dfrac{{a + b\cos x + \operatorname{csin} x}}{{e + f\cos x + g\sin x}}} dx$

Working Rule

(i) Put Numerator $ = I$ (denominator) $ + {\text{m}}$ (derivative of denominator) $ + {\text{n}}$

$a + b\cos x + c\sin x = l(e + f\cos x + g\sin x) + m$

$( - f\sin x + g\cos x) + n$

(ii) Equate coefficients of $\sin x,\cos x$ and constant term n on both sides and find the values of $l,\;{\text{m}},{\text{n}}$.

(iii) The next step is to divide the provided integral into$3$ distinct integrals and evaluate each one separately, i.e.

$\int {\dfrac{{a + b\cos x + \operatorname{csin} x}}{{e + f\cos x + g\sin x}}} dx$

$l\int 1 {\text{dx}} + {\text{m}}\int {\dfrac{{ - {\text{f}}\sin {\text{x}} + {\text{g}}\cos {\text{x}}}}{{{\text{e}} + {\text{f}}\cos x + {\text{g}}\sin x}}} {\text{dx}} + $

${\text{n}}\int {\dfrac{{{\text{dx}}}}{{{\text{e}} + {\text{f}}\cos x + {\text{g}}\sin x}}} $

$ = {\text{Lx}} + {\text{m}}\log |{\text{e}} + {\text{f}}\cos {\text{x}} + {\text{g}}\sin {\text{x}}| + {\text{n}}\int {\dfrac{{d{\text{x}}}}{{{\text{e}} + {\text{f}}\cos {\text{x}} + {\text{g}}\sin x}}} {\text{dx}}$

(iv) Substitute the values of $l,\;{\text{m}},{\text{n}}$ found in Step (ii).

5.5 Integrals of the Form $\int {{{\sin }^{\text{m}}}} x{\cos ^{\text{n}}}{\text{xdx}}$

Working Rule

(i) If the power of $\sin x$ is an odd positive integer, put $\cos x = t$.

(ii) If the power of $\cos {\text{x}}$ is an odd positive integer, put $\sin {\text{x}} = {\text{t}}$.

(iii) If the power of $\sin {\text{x}}$ and $\cos {\text{x}}$ are both odd positive integers, put $\sin x = t$ or $\cos x = t$

(iv) If the power of $\sin x$ and $\cos x$ are both even positive integers, use De' Moivre's theorem as follows:

Let, $\quad \cos x + i\sin x = z$. Then $\cos x - i\sin x = {z^{ - 1}}$

Adding these, we get $z + \dfrac{1}{z} = 2\cos x$ and $z - \dfrac{1}{z} = 2i\sin x$

By De Moivre's theorem, we have

${z^n} + \dfrac{1}{{{z^n}}} = 2\cos nx$ and ${z^n} - \dfrac{1}{{{z^n}}} = 2i{\sin ^n}x$

$\therefore $

${\sin ^m}x{\cos ^n}x = \dfrac{1}{{{{(2i)}^m}}} \cdot \dfrac{1}{{{2^n}}}{\left( {z + \dfrac{1}{z}} \right)^n}{\left( {z - \dfrac{1}{z}} \right)^m}$

$ = \dfrac{1}{{{2^{m + n}}}} \cdot \dfrac{1}{{{i^m}}}{\left( {z + \dfrac{1}{z}} \right)^n}{\left( {z - \dfrac{1}{z}} \right)^m}$

Using the Binomial theorem, extend each of the components on the R.H.S. Then, equidistant from the start and end, group the words. As a result, all such pairings may be expressed as the sines or cosines of various angles. Term by term, continue to integrate.

(v) If the sum of powers of $\sin x$ and $\cos x$ is an even negative integer, put $\tan x = z$.

Solved Examples:

Example 1 

 Evaluate: ${\text{ }}\int {\left( {{x^3} + 5{x^2} - 4 + \dfrac{7}{x} + \dfrac{2}{{\sqrt x }}} \right)} dx$

Sol: ${\text{ }}\int {\left( {{x^3} + 5{x^2} - 4 + \dfrac{7}{x} + \dfrac{2}{{\sqrt x }}} \right)} dx$

$ = \int {{x^3}} dx + \int 5 {x^2}dx - \int 4 dx + \int {\dfrac{7}{x}} dx + \int {\dfrac{2}{{\sqrt x }}} dx$

$ = \int {{x^3}} dx + 5 \cdot \int {{x^2}} dx - 4 \cdot \int 1  \cdot dx + 7 \cdot \int {\dfrac{1}{x}} dx + 2 \cdot \int {{x^{ - 1/2}}} dx$

$ = \dfrac{{{x^4}}}{4} + 5 \cdot \dfrac{{{x^3}}}{3} - 4x + 7\log |x| + 2\left( {\dfrac{{{x^{1/2}}}}{{1/2}}} \right) + C$

$ = \dfrac{{{x^4}}}{4} + \dfrac{5}{3}{x^3} - 4x + 7\log |x| + 4\sqrt x  + C$

Example 2

Evaluate: $\int {{{\text{e}}^{{\text{x}}\log a}}}  + {{\text{e}}^{{\text{a}}\log x}} + {{\text{e}}^{{\text{a}}\log a}}{\text{dx}}$

Sol. We have,

$\int {{{\text{e}}^{{\text{x}}\log a}}}  + {{\text{e}}^{{\text{a}}\log x}} + {{\text{e}}^{{\text{a}}\log a}}{\text{dx}}$

$ = \int {{{\text{e}}^{\log {a^{\text{x}}}}}}  + {{\text{e}}^{\log {x^{\text{a}}}}} + {{\text{e}}^{\log {{\text{a}}^{\text{a}}}}}{\text{dx}}$

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

$ = \int {{a^x}} dx + \int {{x^a}} dx + \int {{a^a}} dx$

$ = \dfrac{{{a^x}}}{{\log a}} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + {a^a} \cdot x + C$

Example 3

Evaluate: $\int {\dfrac{{{x^4}}}{{{x^2} + 1}}} dx$

Sol. $\int {\dfrac{{{x^4}}}{{{x^2} + 1}}} dx$

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

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

Example 4

Evaluate: $\int {\dfrac{{{2^x} + {3^x}}}{{{5^x}}}} dx$

Sol. $\int {\dfrac{{{2^x} + {3^x}}}{{{5^x}}}} dx$

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

$ = \int {\left[ {{{\left( {\dfrac{2}{5}} \right)}^x} + {{\left( {\dfrac{3}{5}} \right)}^x}} \right]} dx = \dfrac{{{{(2/5)}^x}}}{{{{\log }_c}2/5}} + \dfrac{{{{(3/5)}^x}}}{{{{\log }_c}3/5}} + C$

Example 5

Evaluate: $\int {{{\text{x}}^3}} {\sin ^4}{\text{dx}}$

Sol. We have

${\text{I}} = \int {{{\text{x}}^3}} {\sin ^4}{\text{dx}}$

Let ${x^4} = t\quad  \Rightarrow \quad d\left( {{x^4}} \right) = dt$

$ \Rightarrow \quad 4{x^3}dx = dt \Rightarrow \quad dx = \dfrac{1}{{4{x^3}}}dt$

NCERT Class 12 Revision Notes Maths Chapter 7 Solution PDF

You can study together in a group or access the solutions offline for a quick revision of formulas and solutions with the  Revision Notes Class 12 Maths Chapter 7 available in PDF form on the official website of Vedantu. These NCERT Maths Class 12 Chapter 7 Integrals notes can also be printed out so that you can refer to them as per your convenience.

NCERT Solutions Chapter 7 Class 12 Maths Revision Notes 

Definition of Integration - Integration is adding parts to find the whole and is used in mathematics to find areas, volumes, etc. The integration process is mathematically an inverse of differentiation. So if the differential coefficient of a function F(x) is f(x), which means:

d[F(x)]/dx = f(x) then we can call F(x) as the antiderivative or integral of f(x). It is denoted as ∫f(x)dx = F(x)

The notation ∫ is for integration and f(x) is referred to as the integrand. In this equation, x is the variable of integration and dx represents the integration with respect to x.

Indefinite Integral - In Class 12 Maths Revision Notes solution Chapter 7, you would learn that an indefinite integral is a function that takes another function's antiderivative function. Let us clarify this with the example discussed above where:

If d[F(x)]/dx = f(x) then ∫f(x)dx = F(x). We also know that for any arbitrary constant K:

d[F(x) + K]/dx = d[F(x)]/dx + 0 = f(x) . Or, we can say:

 ∫f(x)dx = F(x) + K

This goes to show that both F(x) and F(x) + K are integrals of the same function f(x). Hence, by changing K's value, we can get different integrals of f(x). By this virtue, the integral of f(x) is not a definite value; hence F(x) is called the indefinite integral of f(x).

Properties of Indefinite Integral

• The derivative and the integrand are equal to each other in an indefinite integral.

(∫f(x)dx)’ = f(x)

• If we calculate the indefinite integral of the sum of two functions, it is equal to the sum of the integrals of two functions:

∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx

• If we calculate the indefinite integral of the difference of two functions, it is equal to the difference of the integrals of two functions:

∫[f(x) - g(x)]dx = ∫f(x)dx - ∫g(x)dx

• If there is a constant in the integral, it can be moved outside of the integration.

∫Cf(x)dx = C∫f(x)dx

• If ∫f(x)dx = F(x) + K, then ∫f(ax + y)dx = 1/a F(ax + y) + K

Some Special Integrals

• ∫dx/(x2 + y2) = 1/y tan-1 x/y + K

• ∫dx/(x2 - y2) = 1/2y * log |x - y/x + y| + K

• ∫dx/(y2 - x2) = 1/2y * log |y + x/y - x| + K

• ∫dx/√(y2 - x2) = Sin-1 x/y + K

• ∫dx/√(x2 + y2) = log |x + √(x2 + y2)| + K

Method of Integration by Parts 

Integration by parts is the method of integration of the product of two functions. Let us say g and z are two functions of x, then:

∫(gz)dx = g ∫ zdx - ∫(dg/dx ∫zdx)dx

Which Can Be Explained In Words As Follows:

Integral of the product of 2 functions = first function * integral of the 2nd function - integral of the differential of the first functions * integral of the second function.

NCERT Class 12 Revision Notes Maths Chapter 7 Solution PDF

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