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# Integration by Substitution - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• 58 Questions around this concept.

## Solve by difficulty

Let $x^2 \neq n \pi-1, n \in N$ then

$\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} d x$ is equal to

Find the integral of $\int 4x \sin \left ( x^{2}-1 \right )dx$:

The integral value of $\int \left ( 4x+1 \right )^3dx$:

Integrate the term $sec^{2}ax$ with respect to $x.$

If $f(x)=\int e^x\left(\tan ^{-1} x+\frac{2 x}{\left(1+x^2\right)^2}\right) d x, f(0)=0$, then the value of $f(1)$ is

The integral $\int \frac{\left(x^8-x^2\right) d x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)}$ is equal to:

For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, If $y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$, and $\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$ then $y\left(\frac{\pi}{4}\right)$ is equal to

## Concepts Covered - 1

Integration Using Substitution

The method of substitution is one of the basic methods for calculating indefinite integrals.

Substitution - change of variable

$\\\mathrm{To\;solve\;the\;integrate\;of\;the\;form}\\\\\mathrm{I=\int f\left ( g(x) \right )\cdot g'(x)\;dx,\;}\\\\\mathrm{\;where\;g(x)\;is\;continuously\;differentiable\;function.}\\\mathrm{put\;\;g(x)=t,\;\;g'(x)\;dx=dt}\\\mathrm{After\;substitution,\;we\;get\;\;\int f(t)\;dt.}\\\text{Evalute this integration and substitute back the value of }t.$

Some standard results using susbtitution

$\\\mathrm{1.\;\;\int\frac{f'(x)}{f(x)}\;dx=\log_e\left | f(x) \right |+c}\\\\\mathrm{2.\;\;\int f'(x)\left ( f(x) \right )^n\;dx=\frac{\left ( f(x) \right )^{n+1}}{n+1}+c}$

Integration of the function f(ax + b)

$\\ {\text {If } \int f(x) d x=F(x)+C \text { and } a, b \text { are constants, then }} \\\\ {\int f(a x+b) d x=\frac{1}{a} F(a x+b)+C}$

$\\\mathrm{we\;have,\;I=\int\;f(ax+b)dx}\\\\\mathrm{let\;\;ax+b=t,\;\;then\;\;a\;dx=dt}\\\\\mathrm{\therefore \;\;\;\;I=\int f(ax+b)\;dx}\\\\\mathrm{\;\;\;\;\;\;\;\;\;=\frac{1}{a}\int f(t)\;dt}\\\\\mathrm{\;\;\;\;\;\;\;\;\;=\frac{1}{a}F(t)+c}\\\\\mathrm{\;\;\;\;\;\;\;\;\;=\frac{1}{a}F(ax+b)+c}$

For example:

$\\\mathrm{1.\;\;\int\cos2x\;dx=\frac{1}{2}\sin 2x+c}\\\\\mathrm{2.\;\;\int\frac{1}{x+1}\;dx=\log_e|x+1|+c}\\\\\mathrm{3.\;\;\int e^{2x-3}\;dx=\frac{1}{2}e^{2x-3}+c}$

Also, Integrals of tan x, cot x, sec x, cosec x  all these can be evaluated using the result :

$\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+C$

$\\\mathrm{(i)\;\;\;\;\;\;\int \tan x d x=\int \frac{\sec x \tan x}{\sec x} d x}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\; \quad \int \tan x d x=\log |\sec x|+C}\\\\\\\mathrm{(ii)\;\;\;\;\int \cot x d x=\int \frac{\cos x}{\sin x} d x=\log |\sin x|+C}\\\\\\\mathrm{(iii)\;\;\;\;\int \sec x d x=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} d x=\int \frac{\sec ^{2} x \sec x+\tan x}{\sec x+\tan x} d x}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\; \quad \int \sec x d x=\log |\sec x+\tan x|+C}\\\\\\\mathrm{(iv)\;\;\;\;\int \csc x d x=\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x} d x=\int \frac{\csc ^{2} x-\csc x \cot x}{\csc x-\cot x} d x}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\; \quad \int \csc x d x=\log |\csc x-\cot x|+C}$

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Integration Using Substitution

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## Books

### Reference Books

#### Integration Using Substitution

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.4

Line : 39