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.

Integration by Substitution

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

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

Integration of Type $\int\left(\sin ^m x \cdot \cos ^n x\right) d x$

Substitution

। Put $t=\cos x$
II Put $t=\sin x$
III Put $t=\sin x$ or $t=\cos x$
iV If $(m+n)<0$ put $t=\tan x$ or, If $(m+n)>0$ use DeMoivre's Theorem.

$
\left(\sin ^m x\right)\left(\cos ^n x\right) \therefore \int\left(\sin ^m x \cos ^n x\right) d x
$

Case

I Odd Even

II Even Odd

III Odd Odd

IV Even Even

Working rule:

1. If one of them is odd, then substitute the term with even power as t.

2. If both are odd, substitute either of them as t.

3. If both are even, use trigonometric identities of multiple angles

4. If m and n are rational numbers and (m + n -2)/2 is a negative integer, then substitute cot x = p or tan x = p for different cases of a type of indefinite integration

Integrals of the Form $\int x^m\left(a+b x^n\right)^p d x$

Working Rule:

Case I: If $P \in N$, expand using binomial and integrate.
Case II : If $P \in I^{-}$(ie, negative integer), write $x=t^k$ where $k$ is the LCM of $m$ and $n$.

Case III : If $\frac{m+1}{n}$ is an integer and $P \leftrightarrow$ fraction, put $\left(a+b x^n\right)=t^k$ where $k$ is denominator of the fraction $P$.

Case IV : If $\left(\frac{m+1}{n}+P\right)$ is an integer and $P \in$ fraction, $\operatorname{put}\left(a+b x^n\right)=t^k x^n$, where $k$ is denominator of the fraction $P$

Integration of the form $\int f\left(x, \sqrt{a x^2+b x+c}\right) d x$ can be solved using one of the Euler's substitutions.
(i) $\sqrt{a x^2+b x+c}=t \pm x \sqrt{a}$, if $a>0$
(ii) $\sqrt{a x^2+b x+c}=t x+\sqrt{c}$, if $c>0$
(iii) $\sqrt{a x^2+b x+c}=(x-\alpha) t$, If $\alpha$ is real root of $a x^2+b x+c$

Illustration

The value of the integral, $I=\int \frac{\mathrm{dx}}{1+\sqrt{\mathrm{x}^2+2 \mathrm{x}+2}}$

Here a = 1 > 0, therefore make the substitution $\sqrt{x^2+2 x+2}=t-x$

Squaring both sides of this equality and reducing the similar terms, we get

$\left\{\begin{array}{l}x^2+2 x+2=(t-x)^2 \\ x^2+2 x+2=t^2-2 t x+x^2 \\ 2 x+2 t x=t^2-2\end{array}\right.$

$
\begin{aligned}
\Rightarrow & x=\frac{t^2-2}{2(1+t)} \Rightarrow d x=\frac{t^2+2 t+2}{2(1+t)^2} d t \\
& 1+\sqrt{\mathrm{x}^2+2 \mathrm{x}+2}=1+\mathrm{t}-\frac{\mathrm{t}^2-2}{2(1+\mathrm{t})}=\frac{\mathrm{t}^2+4 \mathrm{t}+4}{2(1+\mathrm{t})^2}
\end{aligned}
$
Substituting into the integral, we get

$
\mathrm{I}=\int \frac{2(1+\mathrm{t})\left(\mathrm{t}^2+2 \mathrm{t}+2\right)}{\left(\mathrm{t}^2+4 \mathrm{t}+4\right) 2(1+\mathrm{t})^2} \mathrm{dt}=\int \frac{\left(\mathrm{t}^2+2 \mathrm{t}+2\right) \mathrm{dt}}{(1+\mathrm{t})(1+2)^2}
$

Using the partial fraction,

$
\frac{t^2+2 t+2}{(t+1)(t+2)^2}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2)^2}
$

we get $\mathrm{A}=1, \mathrm{~B}=0$ and $\mathrm{C}=-2$

$
\frac{t^2+2 t+2}{(t+1)(t+2)^2}=\frac{1}{t+1}-\frac{2}{(t+2)^2}
$
Hence,

$
\begin{aligned}
\int \frac{t^2+2 t+2}{(1+t)(1+2)^2} d t & =\int \frac{d t}{t+1}-2 \int \frac{d t}{(t+2)^2} \\
& =\ln |t+1|+\frac{2}{t+2}+C
\end{aligned}
$

Now, put the value of t in terms of x

$I=\ln \left(x+1+\sqrt{x^2+2 x+2}\right)+\frac{2}{x+2+\sqrt{x^2+2 x+2}}+C$

Solved Examples Based On Integration by Substitution:

Example 1: Evaluate $\int(\sin x)^4(\cos x)^{-6} d x$

1) $\frac{\sin ^5 x(\cos x)^{-5}}{5}+C$
2) $\frac{\sin ^5 x(\cos x)^{-7}}{35}+C$
3) $\frac{\sin ^5 x(\cos x)^{-6}}{5}+C$

4) $\frac{(\sin x)^5}{5}+C$

Solution

As we have learned

The special type of indefinite integration -

The integral of the form $\left(\sin ^m x\right)\left(\cos ^n x\right) \therefore \int\left(\sin ^m x \cos ^n x\right) d x$

- wherein

Where $m, n>0$

In some cases, it may be $m, n<0$
$
\int \frac{\sin ^4 x}{\cos ^4 x} \times \frac{1}{\cos ^2 x} d x=\int \tan ^4 x \sec ^2 x d x=\frac{\tan ^5 x}{5}+C
$
Hence, the answer is the option 1.

Example 2: Find the integral

$
(\sin x)^{5 / 2}\left(\cos ^3 x\right) d x
$

1) $2 / 7(\sin x)^{7 / 2}+C$
2) $2 / 7(\sin x)^{7 / 2} \cos x+C$

$
\begin{aligned}
& \text { 3) } \frac{2}{7}(\sin x)^{7 / 2}-\frac{2}{11}(\sin x)^{11 / 2}+C \\
& \text { 4) } 2 / 7(\sin x)^{7 / 2}-2 / 5(\sin x)^{5 / 2}+C
\end{aligned}
$

Solution

As we have learned

Special type of indefinite integration -

Integral of the form $\left(\sin ^m x\right)\left(\cos ^n x\right) \therefore \int\left(\sin ^m x \cos ^n x\right) d x$

- wherein

Where $m, n>0$

In some case it may be $m, n<0$
$\int(\sin x)^{5 / 2}\left(\cos ^2 x\right) \cos x d x$
$\Rightarrow \int(\sin x)^{5 / 2}\left(1-\sin ^2 x\right) \cos x d x$
$\Rightarrow \int\left((\sin x)^{5 / 2}-\left(\sin ^{9 / 2} x\right)\right) \cos x d x$
$\Rightarrow \frac{2}{7}(\sin x)^{7 / 2}-\frac{2}{11}(\sin x)^{11 / 2}+C$

Hence, the answer is the option 3.

Example 3: $\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{(\sin x)^7}+C$, then $f(x)=$

1) sin x

2) cos x

3) tan x

4) cot x

Solution

As we learned

Different cases of a type of indefinite integration -

$\left(\sin ^m x\right)\left(\cos ^n x\right) \therefore \int\left(\sin ^m x \cos ^n x\right) d x$

Case

I Odd Even

II Even Odd

III Odd Odd

IV Even Even

- wherein

Substitution

I Put $t=\cos n$

II Put $t=\sin x$

III Put$t=\sin x$ or $t=\cos x$

iV If $(m+n)<0$ put$t=\tan x$ or, If $(m+n)>0$ use DeMoivre's Theorem.

$\int_{I_1-I_2} \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\int\left(\frac{\sec ^2 x}{\sin ^7 x}-\frac{7}{\sin ^7 x}\right) d x=\int \frac{\sec ^2 x}{\sin ^7 x} d x-\int \frac{7}{\sin ^7 x} d x=$

Now $I_1=\int \frac{\sec ^2 x}{\sin ^7 x} d x=\frac{\tan x}{\sin ^7 x}+7 \int \frac{\tan x \cdot \cos x}{\sin ^8 x} d x=\frac{\tan x}{\sin ^7 x}+I_2$$\therefore I_1-I_2=\frac{\tan x}{\sin ^7 x}+C \therefore f(x)=\tan x$

Hence, the answer is the option (3).

Example 4: Evaluate $\int \sin ^2 x \cos ^4 x d x$.

1) $\frac{\sin 6 x}{192}-\frac{\sin 4 x}{64}-\frac{1}{64} \sin 2 x+\frac{x}{16}+c$

2) $-\frac{\sin 6 x}{192}-\frac{\sin 4 x}{64}+\frac{1}{64} \sin 2 x+\frac{x}{16}+c$

3) $-\frac{\sin 6 x}{192}-\frac{\sin 4 x}{64}+\frac{1}{64} \sin 2 x-\frac{x}{16}+c$

4) $\frac{\sin 6 x}{192}-\frac{\sin 4 x}{64}+\frac{1}{64} \sin 2 x-\frac{x}{16}+c$

Solution:

We have $\int \sin ^2 x \cos ^4 x d x$ $=\int\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{\cos 2 x+1}{2}\right)^2 d x$

$\begin{aligned} & =\frac{1}{8} \int(1-\cos 2 x)\left(\cos ^2 2 x+2 \cos 2 x+1\right) d x \\ & =\frac{1}{8} \int\left(\frac{\cos 4 x+1}{2}\right) d x+\int \frac{2 \cos 2 x d x}{8}+\int \frac{d x}{8}-\frac{1}{8} \int \cos ^3 2 x d x\end{aligned}$

$-\frac{1}{4} \int \cos ^2 2 x d x-\frac{1}{8} \int \cos 2 x d x$

$I=\frac{1}{16} \frac{\sin 4 x}{4}+\frac{x}{16}+\frac{\sin 2 x}{8}+\frac{x}{8}-\frac{1}{8} \int \frac{\cos 6 x+3 \cos 2 x}{4} d x$

$-\frac{1}{8}\left[(\cos 4 x+1) d x-\frac{1}{16} \sin 2 x\right]$

$\begin{aligned} & \quad=\frac{1}{64} \sin 4 x+\frac{1}{16} \sin 2 x+\frac{3 x}{16}-\frac{1}{32}\left[\frac{\sin 6 x}{6}+\frac{3 \sin 2 x}{2}\right]- \\ & \frac{1}{32} \sin 4 x-\frac{x}{8}\end{aligned}$

$=-\frac{\sin 6 x}{192}-\frac{\sin 4 x}{64}+\frac{1}{64} \sin 2 x+\frac{x}{16}+c$

Hence, the answer is the option (2).

Example 5: The value of the integral $\int \frac{\mathrm{dx}}{x \sqrt{x^2+x-2}}$ is

1) $\sqrt{2} \arctan \left(\frac{\sqrt{2} \sqrt{x-1}}{\sqrt{x+2}}\right)+C$
2) $2 \sqrt{2} \arctan \left(\frac{\sqrt{x-1}}{\sqrt{x+2}}\right)+C$
3) $\sqrt{2} \arctan \left(\frac{\sqrt{x+2}}{\sqrt{2} \sqrt{x-1}}\right)+C$
4) $\sqrt{2} \arctan \left(\frac{\sqrt{2} \sqrt{x+2}}{\sqrt{x-1}}\right)+C$

Solution

$
\begin{aligned}
& I=\int \frac{d x}{x \sqrt{x^2+x-2}} \\
& \because x^2+x-2=(x+2)(x-1) \\
& \operatorname{put} \sqrt{x^2+x-2}=\sqrt{(x+2)(x-1)}=(x-1) t
\end{aligned}
$

square both side, we get

$
\begin{aligned}
& (x+2)=(x-1) t^2 \\
& 2+t^2=x\left(t^2-1\right) \\
& \mathrm{x}=\frac{2+t^2}{\mathrm{t}^2-1} \Rightarrow \mathrm{dx}=-\frac{6 \mathrm{t}}{\left(\mathrm{t}^2-1\right)^2} \mathrm{dt}
\end{aligned}
$
Substituting into the integral, we obtain

$
\Rightarrow \int \frac{-\frac{6 \mathrm{t}}{\left(\mathrm{t}^2-1\right)^2} \mathrm{dt}}{\left(\frac{2+\mathrm{t}^2}{\mathrm{t}^2-1}\right)\left(\frac{2+\mathrm{t}^2}{\mathrm{t}^2-1}-1\right) \mathrm{t}}=-\int \frac{2}{\mathrm{t}^2+2} \mathrm{dt}
$

$
-\int \frac{2}{t^2+2} \mathrm{dt}=-\sqrt{2} \arctan \left(\frac{\mathrm{t}}{\sqrt{2}}\right)+C
$
Now, put the value of $t$

$
=\sqrt{2} \arctan \left(\frac{\sqrt{2} \sqrt{x-1}}{\sqrt{x+2}}\right)+C
$

Hence, the answer is the option 1.

Summary

Integration by substitution is a powerful technique used in calculus to simplify the process of finding integrals. It is an important concept in mathematics. It provides a deeper understanding of mathematical ideas paramount for later developments in many scientific and engineering disciplines.