Approach that I have tried:
$u = x$ $\to$ $u' = 1$
$v'= \cot x\to v = \ln(\sin x)$
$$\int x\cot^2x\,dx=x\ln(\sin x)-\int\ln(\sin x)\,dx$$
And from there, I have a problem, because solving for integral of ln(sinx) gives complicated results with I doubt give correct final answer. I would appreciate your help.
$$ \begin{aligned} \int x\cot ^2 x d x & =\int x\left(\csc ^2 x-1\right) d x \\ & =-\int x d(\cot x)-\frac{x^2}{2} \\ & =-x \cot x+\int \cot x d x-\frac{x^2}{2} \\ & =-x \cot x+\int \frac{d(\sin x)}{\sin x}-\frac{x^2}{2} \\ & =-x \cot x+\ln |\sin x|-\frac{x^2}{2}+C \end{aligned} $$
Just see that the problem have been solved, so just offer another possible solution=)
Let $u = \tan x$, we then have
$$\int x\cot^2x\,dx=\int \left(\frac{\arctan(u)}{u^2}-\frac{\arctan(u)}{u^2+1} \right) \, du$$ For the first part, apply integration by parts (let $v = \frac{1}{u}$); for the second part, let $s = \arctan(u)$.
