# Examples On Integration By Substitution Set-2

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## Examples on integration

Example - 13

Evaluate \begin{align}\int {\frac{1}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}}dx} \end{align} .

Solution:

$I = \int {\frac{1}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}}dx}$

$= \int {\frac{1}{{{{\left( {{{\left( {x + 1} \right)}^2} + 1} \right)}^2}}}dx}$

The denominator can be reduced by the substitution

\begin{align}&\qquad\qquad\quad x + 1 = \tan \theta \\&\Rightarrow \quad {\left( {x + 1} \right)^2} + 1 = {\tan ^2}\theta + 1 = {\sec ^2}\theta \end{align}

Also, $dx = {\sec ^2}\theta d\theta$

\begin{align} &\;\Rightarrow \quad I = \int {\frac{{{{\sec }^2}\theta }}{{{{\sec }^4}\theta }}d\theta } \\&\qquad\quad= \int {{{\cos }^2}\theta d\theta } \\& \qquad\quad= \frac{1}{2}\int {\left( {1 + \cos 2\theta } \right)d\theta } \\ &\qquad\quad= \frac{1}{2}\left( {\theta + \frac{{\sin 2\theta }}{2}} \right) + C\\ &\qquad\quad= \frac{1}{2}\left( {\theta + \sin \theta \cos \theta } \right) + C\end{align}

To express the integral in terms of x, we use

\begin{align}&\qquad\;\; x+1=\tan \theta \\ & \quad\;\;\Rightarrow \quad \theta ={{\tan }^{-1}}\left( x+1 \right) \\ &\quad\; \Rightarrow \quad \sin \theta =\frac{\left( x+1 \right)}{\sqrt{1+{{\left( x+1 \right)}^{2}}}} \\ & and\qquad \cos \theta =\frac{1}{\sqrt{1+{{\left( x+1 \right)}^{2}}}} \\ &\qquad\;\; \Rightarrow \quad I=\frac{1}{2}\left\{ {{\tan }^{-1}}\left( x+1 \right)+\frac{x+1}{{{x}^{2}}+2x+2} \right\}+C \\ \end{align}

Example - 14

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

Solution: A slight thought on the form of this expression will hint that a trigonometric substitution might help; recall that both $$\left( {1 - \cos \theta } \right)and\left( {1 + \cos \theta } \right)$$ and can be reduced to single terms. Therefore, we use the substitution

\begin{align}&\qquad \sqrt x = \cos \theta \\ &\;\Rightarrow \quad x = {\cos ^2}\theta \\&\Rightarrow \quad dx = - 2\sin \theta \cos \theta d\theta \end{align}

\begin{align} & I=\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}.\frac{1}{x}dx}=\int{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\frac{1}{{{\cos }^{2}}\theta }.\left( -2\sin \theta \cos \theta d\theta \right)} \\ &\qquad\qquad\qquad\qquad\qquad =-2\int{\sqrt{\frac{2{{\sin }^{2}}\left( {}^{\theta }\!\!\diagup\!\!{}_{2}\; \right)}{2{{\cos }^{2}}\left( {}^{\theta }\!\!\diagup\!\!{}_{2}\; \right)}}\tan \theta d\theta } \\ &\qquad\qquad\qquad\qquad\qquad =-2\int{\tan \left( \frac{\theta }{2} \right).\tan \theta }d\theta \\ &\qquad\qquad\qquad\qquad\qquad =-4\int{\frac{{{\sin }^{2}}\left( {}^{\theta }\!\!\diagup\!\!{}_{2}\; \right)}{\cos \theta }}d\theta \\ &\qquad\qquad\qquad\qquad\qquad =-2\int{\frac{1-\cos \theta }{\cos \theta }}d\theta \\ &\qquad\qquad\qquad\qquad\qquad =-2\int{\left( \sec \theta -1 \right)}d\theta \\ & \qquad\qquad\qquad\qquad\qquad=-2\left\{ \ln \left| \sec \theta +\tan \theta \right|-\theta \right\}+C \\ & \qquad\qquad\qquad\qquad\qquad=-2\ln \left| \frac{1+\sin \theta }{\cos \theta } \right|+2\theta +C \\ &\qquad\qquad\qquad\qquad\qquad =-2\ln \left( \frac{1+\sqrt{1-x}}{\sqrt{x}} \right)+2{{\cos }^{-1}}\sqrt{x}+C \\ \end{align}