Integration by Substitution Method: Definition & Example

Integration by Substitution Method: Definition, Formula

The method of substitution is one of the most powerful and commonly used techniques in integration. It helps convert a complicated integral into a simpler one by changing the variable of integration. This is particularly helpful for composite functions, where direct integration is difficult.

For example, the integral $ \int \sin(3x + 2) , dx $ may look tricky at first glance — but using substitution, we can set $ t = 3x + 2 $, which makes it straightforward to integrate.

Concept of Integration

Let $f$ be a function of $x$ defined on the closed interval $[a, b]$, and let $F$ be another function such that

$\frac{d}{dx}(F(x)) = f(x)$

for all $x$ in the domain of $f$. Then,

$\int_a^b f(x)\, dx = [F(x)]_a^b = F(b) - F(a)$

is called the definite integral of the function $f(x)$ over the interval $[a, b]$, where $a$ is the lower limit and $b$ is the upper limit of integration.

Steps to Evaluate a Definite Integral

1. Find the indefinite integral $\int f(x), dx$ and let the result be $F(x)$.

2. Evaluate $F(b)$ and $F(a)$.

3. The definite integral is then obtained as $\int_a^b f(x)\, dx = F(b) - F(a)$

This process helps find the net area under the curve of $f(x)$ between $x = a$ and $x = b$.

Integration by Substitution (Change of Variable)

When an integral involves a composite function, substitution is often the easiest way to simplify it.

Suppose you have an integral of the form

$I = \int f(g(x)) \cdot g'(x)\, dx$

where $g(x)$ is a continuously differentiable function.
Let $t = g(x)$, so $dt = g'(x), dx$.

After substitution, the integral becomes

$I = \int f(t)\, dt$

Integrate with respect to $t$, then substitute back $t = g(x)$ to get the final answer in terms of $x$.

Standard Results Using Substitution

Some useful formulas derived using integration by substitution are:

1. $\int \frac{f'(x)}{f(x)}, dx = \log_e |f(x)| + C$

2. $\int f'(x),(f(x))^n, dx = \frac{(f(x))^{n+1}}{n+1} + C$

These results are particularly handy for solving complex integrals quickly.

Integration of Functions of the Form $f(ax + b)$

If $\int f(x), dx = F(x) + C$ and $a, b$ are constants, then

$\int f(ax + b)\, dx = \frac{1}{a} F(ax + b) + C$

Derivation

Let $I = \int f(ax + b)\, dx$

Substitute $t = ax + b$, then $dt = a, dx$ or $dx = \frac{dt}{a}$.

Hence, $I = \frac{1}{a} \int f(t)\, dt = \frac{1}{a} F(t) + C = \frac{1}{a} F(ax + b) + C$

Examples of Integration by Substitution

1. $\int \cos 2x, dx = \frac{1}{2} \sin 2x + C$

2. $\int \frac{1}{x+1}, dx = \log_e |x+1| + C$

3. $\int e^{2x - 3}, dx = \frac{1}{2} e^{2x - 3} + C$

Each of these examples uses substitution to make the integral straightforward.

Integration of Trigonometric Functions Using Substitution

Trigonometric functions like $\tan x$, $\cot x$, $\sec x$, and $\csc x$ can also be integrated using the rule

$\int \frac{f'(x)}{f(x)}\, dx = \log |f(x)| + C$

1. Integration of $\tan x$

$\int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx$

Let $f(x) = \cos x$, then $f'(x) = -\sin x$.

So, $\int \tan x\, dx = -\log |\cos x| + C = \log |\sec x| + C$

2. Integration of $\cot x$

$\int \cot x\, dx = \int \frac{\cos x}{\sin x}\, dx = \log |\sin x| + C$

3. Integration of $\sec x$

$\int \sec x\, dx = \log |\sec x + \tan x| + C$

4. Integration of $\csc x$

$\int \csc x\, dx = \log |\csc x - \cot x| + C$

Fundamental Integration Formulae

Here are some of the most important and commonly used formulas:

1. $\int x^n, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

2. $\int \sin x, dx = -\cos x + C$

3. $\int \cos x, dx = \sin x + C$

4. $\int e^x, dx = e^x + C$

5. $\int \frac{1}{x}, dx = \log |x| + C$

If $x$ is replaced by a linear function of $x$, say $(ax + b)$, then:

$\int f(ax + b)\, dx = \frac{F(ax + b)}{a} + C$

Key Points

• Integration by substitution helps simplify composite or nested functions.

• Identify an inner function $g(x)$ and replace it with a new variable $t$.

• After integrating, substitute back the original expression.

• This method is highly effective for trigonometric, exponential, and rational integrals.

• Mastering substitution is essential for understanding advanced techniques like integration by parts and reduction formulas.

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 3: 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 4: 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.

List of Topics Related to Integration by Substitution Method

This section outlines the key topics related to the integration by substitution method, helping you understand its complete scope and related concepts.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by Parts

Application of Inequality in Definite Integration

NCERT Resources

Access all the essential NCERT materials, including notes, solutions, and exemplars, to strengthen your understanding of integration concepts.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Integration by Substitution Method

Strengthen your understanding with practice questions designed to test the application of the substitution method in various integral problems.

Integration By Substitution- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: