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