$\underset{0}{\overset{\pi }{\int }}x\ln \left(Sinx\right)dx=$

$$\begin{gathered} I=\underset{0}{\overset{\pi }{\int }}x\ln \left(Sinx\right)dx \\ =\underset{0}{\overset{\pi }{\int }}\left(\pi -x\right) \ln \left(\sin \left(\pi -x\right)\right)dx \\ =\underset{0}{\overset{\pi }{\int }}\left(\pi -x\right) \ln \sin x dx \\ I=\pi \underset{0}{\overset{\pi }{\int }}\ln \sin x dx-I \\ 2I=\pi \underset{0}{\overset{\pi }{\int }}\ln \sin x dx \\ I=\frac{\pi }{2}\underset{0}{\overset{\pi }{\int }}\ln \sin x dx \\ I=\frac{\pi }{2}\left(2\underset{0}{\overset{\pi /2}{\int }} \ln \sin x dx\right) \\ \left[\underset{0}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a/2}{\int }}f\left(x\right)dx if f\left(a-x\right)=f\left(x\right)\right] \\ =\pi \underset{0}{\overset{\pi /2}{\int }}\ln \sin x dx \\ =\pi \left(\frac{\pi }{2}\ln 2\right) \\ =\frac{-{\pi }^2}{2}\ln 2 \end{gathered}$$