Question

${\int }_0^{\infty }{\left({\cot }^{-1}x\right)}^{2}dx=\frac{\pi }{k}\log 2$. Find the value of $k$.

Answer 1

Let $I={\int }_0^{\infty }{\left({\cot }^{-1}x\right)}^{2}dx$
Put $x=\cot \theta$ and adjust the limits.
$\therefore I={\int }_{\pi /2}^0{\theta }^2(-co{\sec }^2\theta )d\theta$
$={\int }_0^{\pi /2}{\theta }^{2}co{\sec }^2\theta d\theta$
$={\left[{\theta }^2(-\cot \theta )\right]}_0^{\pi /2}+2{\int }_0^{\pi /2}\theta \cot \theta d\theta$
$=0+2.\frac{\pi }{2}\log 2=\pi \log 2,$