Integration by Substitution Method: Definition & Example

Integration by Substitution

The method of substitution is one of the basic methods for calculating indefinite integrals. This technique transforms a complex integral into a simpler one by changing the variable of integration. It is especially useful for integrals involving composite functions where a direct integration approach is difficult.

Let f be a function of x defined on the closed interval $[\mathrm{a}, \mathrm{b}]$ and F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all $x$ in the domain of $f$, then

$
\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)
$

is called the definite integral of the function $f(x)$ over the interval $[a, b]$, where $a$ is called the lower limit of the integral and $b$ is called the upper limit of the integral.
Working Rule to evaluate definite Integral $\int_a^b f(x) d x$
1. First find the indefinite integration $\int f(x) d x$ and suppose the result be $\mathrm{F}(\mathrm{x})$
2. Next find the $F(b)$ and $F(a)$
3. And, finally value of definite integral is obtained by subtracting $\mathrm{F}(\mathrm{a})$ from $\mathrm{F}(\mathrm{b})$.
Thus, $\quad \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=[\mathrm{F}(\mathrm{x})]_{\mathrm{a}}^{\mathrm{b}}=\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})$.

Substitution - change of variable
To solve the integrate of the form

$I=\int f(g(x)) \cdot g^{\prime}(x) d x$

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

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

Integration of the function f(ax + b)

Integration of the function $f(a x+b)$
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 $\mathrm{adx}=\mathrm{d} t$

$\begin{aligned}
\therefore \quad \mathrm{I} & =\int \mathrm{f}(\mathrm{ax}+\mathrm{b}) \mathrm{dx} \\
& =\frac{1}{\mathrm{a}} \int \mathrm{f}(\mathrm{t}) \mathrm{dt} \\
& =\frac{1}{\mathrm{a}} \mathrm{F}(\mathrm{t})+\mathrm{c} \\
& =\frac{1}{\mathrm{a}} \mathrm{F}(\mathrm{ax}+\mathrm{b})+\mathrm{c}
\end{aligned}$

For example:
1. $\int \cos 2 x d x=\frac{1}{2} \sin 2 x+c$
2. $\int \frac{1}{x+1} d x=\log _e|x+1|+c$
3. $\int e^{2 x-3} d x=\frac{1}{2} e^{2 x-3}+c$

Also, Integrals of $\tan x, \cot x, \sec x, \operatorname{cosec} x$ all these can be evaluated using the result :

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

(I) $\begin{array}{ll}
\int \tan x d x= & \int \frac{\sec x \tan x}{\sec x} d x \\
\Rightarrow \quad & \int \tan x d x=\log |\sec x|+C
\end{array}$

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

$\Rightarrow \quad \int \csc x d x=\log |\csc x-\cot x|+C$

Fundamental formulae such as $\int x^n d x=\frac{x^{n+1}}{n+1}, \int \sin x d x=-\cos x$, ,... and so on
It $x$ is replaced by a LINEAR FUNCTION of $x \Rightarrow(a x+b)$ form then,

$\int f(a x+b) d x=\frac{F(a x+b)}{\frac{\mathrm{d}}{\mathrm{d} x}(a x+b)}+c$

Recommended Video Based on Integration by Substitution

Solved Examples Based On Integration by Substitution:

Example 1: If $\int \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}}=(\tan x)^A+C(\tan x)^B+k$, where $k$ is a constant of integration, then $A+B+C$ equals:
1) $\frac{21}{5}$
2) $\frac{16}{5}$
з) $\frac{7}{10}$
4) $\frac{27}{10}$

Solution

As learnt in the concept of Integration by substitution -

$\begin{aligned}
& \int \frac{d x}{\cos ^3 x \sqrt{2 \sin x \cos x \times 2}} \\
& =\frac{1}{2} \int \frac{\sec ^4 x d x}{\sqrt{\tan x}} \\
& =\frac{1}{2} \int \frac{\left(1+\tan ^2 x\right) \sec ^2 x d x}{\sqrt{\tan x}} \\
& =\frac{1}{2} \int(\tan x)^{\frac{-1}{2}} \sec ^2 x d x+\frac{1}{2} \int(\tan x)^{\frac{3}{2}} \sec ^2 x d x \\
& =\frac{1}{2} \frac{(\tan x)^{\frac{1}{2}}}{\frac{1}{2}}+\frac{1}{2} \frac{(\tan x)^{\frac{5}{2}}}{\frac{5}{2}}+C \\
& A=\frac{1}{2} ; B=\frac{5}{2} ; C=\frac{1}{5} \\
& A+B+C=3+\frac{1}{5} \\
& =\frac{16}{5}
\end{aligned}$

Hence, the answer is the option 2.

Example 2: The integral $\int \frac{d x}{(1+\sqrt{x}) \sqrt{x-x^2}}$ is equal to: (where C is a constant of integration.)
1) $-2 \sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C$
2) $-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$
3) $-\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$
4) $\sqrt[2]{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C$

Solution

$\begin{aligned} & \sin x=t \Rightarrow \cos x d x=d t \\ & \int \frac{d t}{t^2\left(1+t^6\right)^{2 / 3}}=\int \frac{d t}{t^3 t^4\left(\frac{1}{t^t}+1\right)^{2 / 3}} \\ & u=\frac{1}{t^6}+1 \Rightarrow d u=-\frac{6}{t^7} d t \\ & \frac{d u}{-6}=\frac{d t}{t^7} \\ & =\int \frac{d u}{-6 u^{2 / 3}}=-\frac{1}{2} u^{1 / 3}+C \\ & =-\frac{1}{2}\left(\frac{1}{t^6}+1\right)^{1 / 3}+C \\ & =-\frac{1}{2}\left(\frac{\left(1+\sin ^6 x\right)^{1 / 3}}{\sin ^2 x}\right) \\ & f(x)=-\frac{1}{2} \frac{1}{\sin ^2 x} \quad \lambda=3 \\ & \lambda f(\pi / 3)=-2\end{aligned}$

Hence, the answer is the option 4.

Example 4: The integral $\int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}}$ is equal to: (where C ia a constant of integration)
1) $-\left(\frac{x-3}{x+4}\right)^{-1 / 7}+C$
2) $\frac{1}{2}\left(\frac{x-3}{x+4}\right)^{3 / 7}+C$
3) $\left(\frac{x-3}{x+4}\right)^{1 / 7}+C$
4) $-\frac{1}{13}\left(\frac{x-3}{x+4}\right)^{-13 / 7}+C$

Solution

$\int\left(\frac{x-3}{x+4}\right)^{\frac{-6}{7}} \frac{1}{(x+4)^2} d x$

Let $\frac{x-3}{x+4}=t^7$

$\begin{aligned}
& \frac{7}{(x+4)^2} d x=7 t^6 d t \\
& \int t^{-6} t^6 d t=t+c \\
& \left(\frac{x-3}{x+4}\right)^{\frac{1}{4}}+c
\end{aligned}$

Hence, the answer is the option 3.

Example 5: The integral $\int \frac{e^{3 \log _e 2 x}+5 e^{2 \log _e 2 x}}{e^{4 \log _e x}+5 e^{3 \log _c x}-7 e^{2 \log _e x}} d x, x>0 \quad$ is equal to : (where c is a constant of integration)
1) $\log _c\left|x^2+5 x-7\right|+c$
2) $4 \log _e\left|x^2+5 x-7\right|+c$
3) $\frac{1}{4} \log _e\left|x^2+5 x-7\right|$
4) $\log _e \sqrt{x^2+5 x-7}+c$

Solution

$\begin{aligned} & e^{\log a^x}=x \log a \\ & e^{\log _x}=x \\ & \text { Let } \mathrm{I}=\int \frac{\mathrm{e}^{3 \log _e 2 \mathrm{x}}+5 \mathrm{e}^{2 \log _e 2 \mathrm{x}}}{\mathrm{e}^{4 \log _e \mathrm{x}}+5 \mathrm{e}^{3 \log _{\mathrm{e}} \mathrm{x}}-7 \mathrm{e}^{2 \log _{\mathrm{e}} \mathrm{x}}} \mathrm{dx}, \mathrm{x}>0 \\ & \mathrm{I}=\int \frac{\mathrm{e}^{\log _e(2 x)^3}+5 \mathrm{e}^{\log _e(2 x)^2}}{\mathrm{e}^{\log _e \mathrm{x}^4}+5 \mathrm{e}^{\log _e \mathrm{x}^3}-7 \mathrm{e}^{\log _e \mathrm{x}^2}} \mathrm{dx} \\ & I=\int \frac{(2 x)^3+5(2 x)^2}{x^4+5 x^3-7 x^2} d x=\int \frac{4 x^2(2 x+5)}{x^2\left(x^2+5 x-7\right)} d x \\ & \text { put } x^2+5 x-7=t \Rightarrow(2 x+5) d x=d t \\ & I=4 \int \frac{d t}{t}=4 \ln |t|=4 \ln \left|x^2+5 x-7\right|\end{aligned}$

Hence, the answer is the option 2.