Integration by Substitution - Practice Questions & MCQ

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

Substitution - change of variable

To solve the integrate of the form
$
I=\int f(g(x)) \cdot g^{\prime}(x) d x
$
where $g(x)$ is continuously differentiable function.
put$\mathrm{g}(\mathrm{x})=\mathrm{t},\quad\mathrm{g}^{\prime}(\mathrm{x}) \mathrm{dx}=\mathrm{dt}$
After substitution, we get $\int \mathrm{f}(\mathrm{t}) \mathrm{dt}$.
Evalute this integration and substitute back the value of $t$.

Some standard results using substitution

1.$\int\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}\mathrm{dx}=\log _{\mathrm{e}}|\mathrm{f}(\mathrm{x})|+\mathrm{c}$
2.$\int\mathrm{f}^{\prime}(\mathrm{x})(\mathrm{f}(\mathrm{x}))^{\mathrm{n}} \mathrm{dx}=\frac{(\mathrm{f}(\mathrm{x}))^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Integration of the function f(ax + b) 

If $\int f(x) d x=F(x)+C$ and $a, b$ are constants, then
$
\int f(a x+b) d x=\frac{1}{a} F(a x+b)+C
$
we have, $I=\int f(a x+b) d x$
let $\mathrm{ax}+\mathrm{b}=\mathrm{t}$, then $\mathrm{adx}=\mathrm{dt}$
$
\begin{aligned}
\therefore \quad \mathrm{I} & =\int \mathrm{f}(\mathrm{ax}+\mathrm{b}) \mathrm{dx} \\
& =\frac{1}{\mathrm{a}} \int \mathrm{f}(\mathrm{t}) \mathrm{dt} \\
& =\frac{1}{\mathrm{a}} \mathrm{~F}(\mathrm{t})+\mathrm{c} \\
& =\frac{1}{\mathrm{a}} \mathrm{~F}(\mathrm{ax}+\mathrm{b})+\mathrm{c}
\end{aligned}
$

For example:

1. $\int \cos 2 \mathrm{xdx}=\frac{1}{2} \sin 2 \mathrm{x}+\mathrm{c}$
2. $\int \frac{1}{x+1} d x=\log _e|x+1|+c$
3.$\int\mathrm{e}^{2\mathrm{x}-3}\mathrm{dx}=\frac{1}{2}\mathrm{e}^{2 \mathrm{x}-3}+\mathrm{c}$

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

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

(i)
$
\begin{aligned}
\int \tan x d x= & \int \frac{\sec x \tan x}{\sec x} d x \\
\Rightarrow \quad & \int \tan x d x=\log |\sec x|+C
\end{aligned}
$
(ii) $\int \cot x d x=\int \frac{\cos x}{\sin x} d x=\log |\sin x|+C$
(iii)
$
\begin{aligned}
\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 \\
\Rightarrow \quad & \int \sec x d x=\log |\sec x+\tan x|+C
\end{aligned}
$
(iv)
$
\begin{aligned}
& \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 \\
& \Rightarrow \quad \int \csc x d x=\log |\csc x-\cot x|+C
\end{aligned}
$

Integration of $\tan ^m(x)$ or $\cot ^m(x):$

Case 1: $m$ is odd.

Use the identity $\sec ^2(x)=1+\tan ^2(x)$ and substitution $u=\tan (x), d u=\sec ^2(x) d x$.

Case 2: $m$ is even.
Express $\tan ^2(x)$ in terms of $\sec ^2(x)$ using:

$
\tan ^2(x)=\sec ^2(x)-1
$

Expand and integrate.