Integration by Substitution Method: Definition & Example

Integration by substitution is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of integration have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of Integration by substitution. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fourteen questions have been asked on this concept, including two in 2016, one in 2018, two in 2019, three in 2020, four in 2021, one in 2022 and two in 2023.

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.