Integral $\int_0^{\pi/2} x\cot(x)dx$, Differentiation w.r.t. parameter only. Integrate using differentiation w.r.t. parameter only: $\int_0^{\pi/2} x\cot(x)dx$.
We can express this as $\int_0^{\pi/2} x\cdot\frac{\cos(x)}{\sin(x)}dx$.
We can substitute $u=\sin(x)$ to start but I am not sure if that will do us any good.
If we use $x_i$ as a parameter, the answer is of the form. $\lim_{x_i \to 1}I(xi)=\lim_{x_i \to 1}\frac{\pi}{2}\ln(\xi+1)$.
NOTE: Only use differentiation with respect to parameters.

found the way to go...We can want to solve
$$
I=\int_0^{\pi/2}x\cot(x) dx,
$$
so we introduce a parameter $\xi$ by writing
$$
I(\xi)=\int_0^{\pi/2} \frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} dx
$$
and in the limit $\xi \to 1$ we recover I.  Taking a derivative we obtain
$$
I'(\xi)=\frac{d}{d\xi}\int_0^{\pi/2} \frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} dx=\int_0^{\pi/2}\frac{\partial}{\partial \xi} \bigg(\frac{\tan^{-1}(\xi\tan(x))}{\tan(x)} \bigg)dx
$$
Now we take the derivative to obtain
$$
I'(\xi)=\int_0^{\pi/2} \frac{dx}{\big(\xi\tan(x)\big)^2+1} =\frac{\pi}{2(\xi+1)}.
$$
We now integrate our result wrt $\xi$ and realizing the constant of integration is zero, we obtain
$$
I(\xi)=\frac{\pi}{2}\ln(\xi+1).
$$
Taking the limit as $\xi \to 1$ we obtain
$$
\lim_{\xi \to 1} I(\xi)=\lim_{\xi \to 1} \frac{\pi}{2}\ln(\xi+1)=\frac{\pi \ln(2)}{2}.
$$
Thus we have shown that
$$
{\boxed{I=\frac{\pi\ln(2)}{2}}}
$$
