Question

Evaluate the integral using substitution:
${\int }_{-1}^1\frac{dx}{x^2+2x+5}$

Answer 1

${\int }_{-1}^1\frac{dx}{x^2+2x+5}$

$={\int }_{-1}^1\frac{dx}{x^2+2x+1+4}$

$={\int }_{-1}^1\frac{dx}{(x+1{)}^2+4}$

Let $x+1=2\tan \theta$

$\Rightarrow \frac{dx}{d\theta }=2{\sec }^2\theta$

$\Rightarrow dx=2{\sec }^2\theta d\theta$

The new limits are

When $x=-1$

$\Rightarrow 2\tan \theta =-1+1\Rightarrow \tan \theta =0\Rightarrow \theta =0$

When $x=1$

$\Rightarrow 1\tan \theta =1+1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4}$

${\int }_{-1}^1\frac{dx}{(x+1{)}^2+4}$

$={\int }_0^{\frac{\pi }{4}}\frac{2{\sec }^2\theta d\theta }{4{\tan }^2\theta +4}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^2\theta d\theta }{{\tan }^2\theta +1}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^2\theta d\theta }{{\sec }^2\theta +1}$

​​​​​​​$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}d\theta$

$=\frac{1}{2}\left[\theta \right]\frac{\pi }{4}=\frac{1}{2}[\frac{\pi }{4}-0]$

​​​​​​​$=\frac{\pi }{8}$