Question

If $\underset{0}{\overset{1}{\int }}\frac{1}{(1+x)(2+x)\sqrt{x(1-x)}}dx=\frac{k\pi }{\sqrt{6}(\sqrt{3}+1)}$, then the value of $k$ is

Answer 1

$I=\underset{0}{\overset{1}{\int }}\frac{1}{(1+x)(2+x)\sqrt{x(1-x)}}dx$
Let $x={\sin }^2\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta$
So ,$I=\underset{0}{\overset{\pi /2}{\int }}\frac{1}{(1+{\sin }^2\theta )(2+{\sin }^2\theta )}d\theta =2\underset{0}{\overset{\pi /2}{\int }}(\frac{1}{1+{\sin }^2\theta }-\frac{1}{2+{\sin }^2\theta })d\theta =2\underset{0}{\overset{\pi /2}{\int }}(\frac{1}{2-{\cos }^2\theta }-\frac{1}{3-{\cos }^2\theta })d\theta =2\underset{0}{\overset{\pi /2}{\int }}(\frac{{\sec }^2\theta }{2{\sec }^2\theta -1}-\frac{{\sec }^2\theta }{3{\sec }^2\theta -1})d\theta =2\underset{0}{\overset{\pi /2}{\int }}(\frac{{\sec }^2\theta }{2{\tan }^2\theta +1}-\frac{{\sec }^2\theta }{3{\tan }^2\theta +2})d\theta$
Now let $t=\tan \theta \Rightarrow dt={\sec }^2\theta d\theta$
$\therefore I=2\underset{0}{\overset{\infty }{\int }}(\frac{1}{1+2t^2}-\frac{1}{2+3t^2})dt=2{[\frac{1}{\sqrt{2}}{\tan }^{-1}\sqrt{2}t-\frac{1}{\sqrt{6}}{\tan }^{-1}\frac{\sqrt{3}}{\sqrt{2}}t]}_0^{\infty }=2[\frac{\pi }{2\sqrt{2}}-\frac{\pi }{2\sqrt{6}}]=\frac{(\sqrt{3}-1)\pi }{\sqrt{6}}=\frac{2\pi }{\sqrt{6}(\sqrt{3}+1)}$

So, $k=2$