Indirect Substitution in Integral - Practice Questions & MCQ

Let’s go through some illustrations

Illustration 1: Evaluate $\int \frac{3 \mathrm{x}^4+4 \mathrm{x}^3}{\left(\mathrm{x}^4+\mathrm{x}+1\right)^2} d x$
Here, $\quad I=\int \frac{3 x^4+4 x^3}{\left(x^4+x+1\right)^2} d x=\int \frac{x^3(3 x+4)}{x^8\left(1+\frac{1}{x^3}+\frac{1}{x^4}\right)^2} d x$
$
\begin{aligned}
& =\int \frac{\left(\frac{3}{x^4}+\frac{4}{x^5}\right)}{\left(1+\frac{1}{x^3}+\frac{1}{x^4}\right)^2} d x \\
\text { Put } \quad & 1+\frac{1}{x^3}+\frac{1}{x^4}=t \\
\Rightarrow \quad & \left(-\frac{3}{x^4}-\frac{4}{x^5}\right) d x=d t \\
I= & -\int \frac{d t}{t^2}=\frac{1}{t}+c \\
= & \frac{1}{1+\frac{1}{x^3}+\frac{1}{x^4}}+c=\frac{x^4}{x^4+x+1}+c
\end{aligned}
$

Illustration 2: Evaluate $\int\left(x^{3 \mathrm{~m}}+\mathrm{x}^{2 \mathrm{~m}}+\mathrm{x}^{\mathrm{m}}\right)\left(2 \mathrm{x}^{2 \mathrm{~m}}+3 \mathrm{x}^{\mathrm{m}}+6\right)^{1 / \mathrm{m}} \mathrm{dx}, \mathrm{x}>0$ here m is any natural number.

Here,
$
\begin{aligned}
\mathrm{I} & =\int\left(\mathrm{x}^{3 \mathrm{~m}}+\mathrm{x}^{2 \mathrm{~m}}+\mathrm{x}^{\mathrm{m}}\right)\left(2 \mathrm{x}^{2 \mathrm{~m}}+3 \mathrm{x}^{\mathrm{m}}+6\right)^{1 / \mathrm{m}} \mathrm{dx} \\
& =\int\left(x^{3 m}+x^{2 m}+x^m\right) \frac{\left(2 x^{3 m}+3 x^{2 m}+6 x^m\right)^{1 / m}}{x} d x \\
& =\int\left(x^{3 m-1}+x^{2 m-1}+x^{m-1}\right)\left(2 x^{3 m}+3 x^{2 m}+6 x^m\right)^{1 / m} d x
\end{aligned}
$

Put $\quad 2 x^{3 m}+3 x^{2 m}+6 x^m=t$
$
\Rightarrow \quad 6 m\left(x^{3 m-1}+x^{2 m-1}+x^{m-1}\right) d x=d t
$
$\therefore$ Eq. (i) becomes,
$
\begin{aligned}
& I=\int t^{1 / m} \frac{d t}{6 m}=\frac{1}{6 m} \cdot \frac{t^{(1 / m)+1}}{(1 / m)+1}+C \\
& I=\frac{1}{6(m+1)}\left[2 x^{3 m}+3 x^{2 m}+6 x^m\right]^{\frac{m+1}{m}}+C
\end{aligned}
$

Sometimes, to solve integration, it is useful to write the integral as a sum of two related integrals which can be evaluated by making suitable substitutions.

Some examples of algebraic Twins are

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

Integration of the form:

1. $\int f\left(x+\frac{1}{x}\right)\left(1-\frac{1}{x^2}\right) d x$

Put $x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t$
2. $\int f\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right) d x$

Put $x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t$
3. $\int \frac{x^2+1}{x^4+k x^2+1} d x$

Divide numerator and denominator by $x^2$
4. $\int \frac{x^2-1}{x^4+k x^2+1} d x$

Divide numerator and denominator by $x^2$

Some Examples of trigonometric Twins are

$\begin{aligned} & \int \sqrt{\tan x} d x, \quad \int \sqrt{\cot x} d x \\ & \int \frac{1}{\sin ^n x+\cos ^n x} d x, \quad n=4,6 \ldots \\ & \int \frac{ \pm \sin x \pm \cos x}{a \pm b \sin x \cos x} d x\end{aligned}$

Some illustrations to see see how to solve such questions.

Illustration 1: Evaluate $\int \sqrt{\tan x} d x$
$
\begin{aligned}
\text { put } \quad \tan \mathrm{x} & =\mathrm{u}^2 \Rightarrow \sec ^2 \mathrm{xdx}=2 \mathrm{u} d \mathrm{u} \\
\Rightarrow \quad & \mathrm{dx} \\
\Rightarrow & =\frac{2 \mathrm{udu}}{1+\mathrm{u}^4} \\
\therefore \quad & \\
\mathrm{I} & =\int \mathrm{u} \cdot \frac{2 \mathrm{udu}}{1+\mathrm{u}^4}=\int \frac{2 \mathrm{u}^2}{1+\mathrm{u}^4} \mathrm{du} \\
& =\int \frac{\mathrm{u}^2+1}{\mathrm{u}^4+1} \mathrm{du}+\int \frac{\mathrm{u}^2-1}{\mathrm{u}^4+1} \mathrm{du} \\
& =\int \frac{1+1 / \mathrm{u}^2}{\mathrm{u}^2+1 / \mathrm{u}^2} \mathrm{du}+\int \frac{1-1 / \mathrm{u}^2}{\mathrm{u}^2+1 / \mathrm{u}^2} \mathrm{du} \\
& =\int \frac{1+1 / \mathrm{u}^2}{(\mathrm{u}-1 / \mathrm{u})^2+2} \mathrm{du}+\int \frac{1-1 / \mathrm{u}^2}{(\mathrm{u}+1 / \mathrm{u})^2-2} \mathrm{du}
\end{aligned}
$

$
\begin{aligned}
\mathrm{I} & =\int \frac{\mathrm{ds}}{\mathrm{~s}^2+(\sqrt{2})^2}+\int \frac{\mathrm{dr}}{\mathrm{r}^2-(\sqrt{2})^2} \\
{[s=u} & \left.-\frac{1}{u} \text { and } r=u+\frac{1}{u}\right] \\
& =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{s}{\sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \log \left|\frac{r-\sqrt{2}}{r+\sqrt{2}}\right|+c \\
& =\frac{1}{\sqrt{2}}\left[\tan ^{-1}\left(\frac{u-1 / u}{\sqrt{2}}\right)+\frac{1}{2} \log \left(\frac{u+\frac{1}{u}-\sqrt{2}}{u+\frac{1}{u}+\sqrt{2}}\right)\right]+C
\end{aligned}
$
substitute back $u=\sqrt{\tan x}$

Illustration 2: $\quad$ Evaluate $\int \frac{1}{\sin ^4 x+\cos ^4 x} d x$
Here, $\quad I=\int \frac{1}{\sin ^4 x+\cos ^4 x} d x$
Dividing numerator and denominator by $\cos ^4 x$, we get
$
\begin{array}{rlrl} 
& & =\int \frac{\sec ^4 x}{\tan ^4 x+1} d x \\
& & & =\int \frac{\sec ^2 x\left(1+\tan ^2 x\right)}{1+\tan ^4 x} d x \\
\text { Put } \quad \tan & x & =u \quad \Rightarrow \quad \sec ^2 x d x=d u \\
\therefore \quad & & I & =\int \frac{u^2+1}{u^4+1} d u \\
I & & =\int \frac{1+1 / u^2}{u^2+1 / u^2} d u=\int \frac{1+1 / u^2}{(u-1 / u)^2+2} d u
\end{array}
$

Again, put $s=u-\frac{1}{u}$
$
\therefore \quad I=\int \frac{d s}{s^2+2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{s}{\sqrt{2}}\right)+C
$