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Integration by Substitution - Practice Questions & MCQ

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

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  • 58 Questions around this concept.

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QuestionMEDIUM

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 

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

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

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

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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

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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:

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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

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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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Integration by Substitution Current Topic

Integration as an Inverse Process of Differentiation
Integration of Trigonometric Functions
Integration by Substitution
Fundamental Formulae of Indefinite Integration (Inverse Trigonometric Functions)
Integrals of Particular Function
Trigonometric Integrals
Integration By Parts Formula
Integration Using Partial Fraction
Integration of Irrational Algebraic Function
Reduction Formula

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Integration Using Substitution

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.4

Line : 39

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