$\text{ Evaluate }{\int }_0^{\pi } x\log \sin xdx$

$\text{ Let }I={\int }_0^{\pi } x\log \sin xdx$--------(1)

$I={\int }_0^{\pi } (\pi -x)\log \sin (\pi -x)dx={\int }_0^{\pi } (\pi -x)\log \sin xdx$---------(2)

$\text{ Adding equations (1) and (2), we get }2I=\pi {\int }_0^{\pi } \log \sin xdx$

$\text{ or }2I=2\pi {\int }_0^{\pi /2} \log \sin xdx$

$$\begin{gathered} I=\pi {\int }_0^{\pi /2} \log \sin xdx ----\left(3\right) \\ I=\pi {\int }_0^{\pi /2} \log \sin (\frac{\pi }{2}-x)dx \\ =\pi {\int }_0^{\pi /2} \log cosxdx ----\left(4\right) \end{gathered}$$

(3) + (4)

$2I=\pi {\int }_0^{\pi /2} \left(\log \sin x+\mathrm{logcos}x\right)dx$

$(take \mathrm{logsin}x\cos x=\log \frac{2\sin x\cos x}{2})$

$=\pi {\int }_0^{\pi /2} \log \sin 2xdx-\pi {\int }_0^{\pi /2} \log 2dx$

put 2x=t,  dx=$\frac{dt}{2}$  UL=$\pi$, LL=0

$2I$=$\frac{\pi }{2}{\int }_0^{\mathrm{\pi }} \log \sin tdt-\pi \log 2\left(\frac{\pi }{2}\right)$

$2I=\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}} \log \sin \mathrm{tdt}-\frac{{\pi }^2}{2}\log 2$

$2I=I-\frac{{\pi }^2}{2}\log 2$

$I=-\frac{{\pi }^2}{2}\log 2$

$=\frac{{\pi }^2}{2}\log (1/2)$