Question

If $\underset{0}{\overset{\infty }{\int }}\frac{dx}{(1+x^2{)}^4}=\frac{K\pi }{32}$, then the value of $K$ is:

Answer 1

$\text{Put}x=\tan \theta$
$\Rightarrow dx={\sec }^2\theta d\theta$
$\therefore I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta }{(1+{\tan }^2\theta {)}^4}d\theta$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{d\theta }{{\sec }^6\theta }=\underset{0}{\overset{\pi /2}{\int }}{\cos }^6\theta d\theta \{\because \underset{0}{\overset{\pi /2}{\int }}{\cos }^{n}xdx=\frac{(n-1)}{n}\cdot \frac{(n-3)}{(n-2)}\cdots \frac{1}{2}\cdot \frac{\pi }{2},\text{when}n\text{is even}\}$
$=[\frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}]\times \left(\frac{\pi }{2}\right)=\frac{5\pi }{32}$
$\therefore$ On comparing, we have $K=5$