# Examples On Integration By Substitution Set-3

Go back to  'Indefinite Integration'

Example - 15

Evaluate  \begin{align} \int {\frac{1}{{x\sqrt {{x^4} - 1} }}dx}\end{align}.

Solution: One possible way to reduce the term $$\sqrt {{x^4} - 1}$$ is to use a trignometric substitution:

\begin{align}&\qquad\;\;{x^2} = \sec \theta \\ &\Rightarrow \quad{x^4} - 1 = {\sec ^2}\theta - 1 = {\tan ^2}\theta \\&\Rightarrow \quad 2xdx \;\;= \sec \theta \tan \theta d\theta \end{align}

\begin{align}& I=\int{\frac{1}{x\sqrt{{{x}^{4}}-1}}dx=\int{\frac{x}{{{x}^{2}}\sqrt{{{x}^{4}}-1}}dx}} \\ & \text{ }\!\!~\!\!\text{ }\ \,=\frac{1}{2}\int{\frac{\sec \theta \tan \theta }{\sec \theta \tan \theta }d\theta } \\ & \ \,=\frac{1}{2}\int{d\theta } \\ & \ \,=\frac{\theta }{2}+C \\ & \ \,=\frac{1}{2}{{\sec }^{-1}}\left( {{x}^{2}} \right)+C \\ \end{align}

Notice that for the five preceding examples, different substitution have been used in all the five. This shows that there is no hard-and-fast rule to do substitutions; you have to judge the most appropriate substitution by analyzing the expression of the function to be integrated.

Example - 16

Evaluate  \begin{align} \int {\frac{{\tan x}}{{a + b\;{{\tan }^2}x}}dx} \end{align}

Solution: We first reduce this expression to another form involving sin and cos terms:

$I = \int {\frac{{\sin x\cos x}}{{a\;{{\cos }^2}x + b\;{{\sin }^2}x}}dx}$

If you observe the expression for $$I$$ carefully, you will realise that a simple substitution in now possible:

\begin{align}&\qquad\quad a\;{\cos ^2}x + b\;{\sin ^2}x = y \\&\Rightarrow \quad ( - 2a\sin x\cos x + 2b\sin x\cos x)dx = dy\\&\Rightarrow \quad \sin x\cos xdx = \frac{{dy}}{{2b - 2a}}\end{align}

Thus we have obtained the numerator in terms of the derivative of the denominator:

\begin{align}&\;I = \frac{1}{{2(b - a)}}\int {\frac{{dy}}{y}}\\&\,\,\,\, = \frac{{\ln |y|}}{{2(b - a)}} + C\\&\,\,\,\, = \frac{{\ln |a\;{{\cos }^2}x + b\;{{\sin }^2}x|}}{{2(b - a)}} + C\end{align}

Example - 17

Evaluate \begin{align}\int {\frac{1}{{\sqrt {{{\sin }^3}x\;\sin (x + a)} }}dx} \end{align}

Solution: As in the previous example, we again have to modify the expression given to us, so that some substitution is possible. The first step that we could take is expand $$sin (x + a)$$ :

\begin{align}& I=\int{\frac{1}{\sqrt{{{\sin }^{3}}x\sin (x+a)}}dx} \\ & \;=\int{\frac{1}{\sqrt{{{\sin }^{3}}x(\sin x\cos a+\cos x\sin a)}}dx} \\ \end{align}

To proceed further, notice that there is a $${\sin ^3}x$$ term in the denominator. What we now do is take out a common factor of  $$\sin x$$  from the (inner) brackets so that $${\sin ^3}x$$ becomes $${\sin ^4}x$$:

\begin{align}& \ I=\int{\frac{1}{\sqrt{{{\sin }^{4}}x\left( \cos a+\cot x\sin a \right)}}dx} \\ & \ \ =\int{\frac{1}{{{\sin }^{2}}x\sqrt{\cos a+\cot x\sin a}}dx} \\ & \ \ =\int{\frac{\text{cose}{{\text{c}}^{2}}x}{\sqrt{\cos a+\cot x\sin a}}dx} \\ \end{align}

This is a form in which the numerator can be expressed as the derivative of the expression in the denominator.

\begin{align} & \text{Substitute}\qquad \qquad \cos a+\cot x\sin a=t \\ & \qquad \qquad \ \Rightarrow \qquad -\sin a\,\text{cose}{{\text{c}}^{2}}xdx=dt \\ & \qquad \qquad \ \Rightarrow \qquad I=-\frac{1}{\sin a}\int{\frac{dt}{\sqrt{t}}} \\ & \qquad \qquad \qquad \qquad \ =\frac{-2}{\sin a}\sqrt{t}+C \\ & \qquad \qquad \qquad \qquad \ =\frac{-2\sqrt{\cos a+\cot x\sin a}}{\sin a}+C \\ \end{align}