Integral of x cotan (x) dx from 0 to pi/2

Did you try doing this 'under the Integral'?. This one is often done that way.

$xcot(x)=\frac{x}{tan(x)}$

$I(a)=\int_{0}^{\frac{\pi}{2}}\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}dx$

$I'(a)=\frac{d}{dx}\int_{0}^{\frac{\pi}{2}}\frac{ta n^{-1}(a\cdot tan(x))}{tan(x)}dx$

$=\int_{0}^{\frac{\pi}{2}}\frac{\partial}{{\partial }b}\left[\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}\right]dx$

$\int_{0}^{\frac{\pi}{2}}\frac{1}{(a\cdot tan(x))^{2}+1}dx$

$=\frac{\pi}{2(a+1)}$

Now, integrate w.r.t a. Then, let a=1 to finish line.

This is a cool method. Learn about 'differentiating under the integral sign'.
You can probably find it somewhere.