TITLE: Verifing $\int_0^{\pi}x\ln(\sin x)\,dx=-\ln(2){\pi}^2/2$
QUESTION [5 upvotes]: I used all I know to show that $$\int_0^\pi x\ln(\sin x)dx=-\ln(2) \pi^2/2$$ This is my homework but don't know where to start. I appreciate your help.

REPLY [6 votes]: This answer is meant to offer an alternative from the last integral of Sasha's work
By variable change $x = 2u$ we have that 
$$I=\int_{0}^{\frac{\pi}{2}} \ln(\sin (x)) \ dx=$$
$$2\int_{0}^{\frac{\pi}{4}} \ln(\sin (2u)) \ du=$$
$$2\left(\int_{0}^{\frac{\pi}{4}} \ln (2) \ du + \int_{0}^{\frac{\pi}{4}} \ln(\sin (u)) \ du +\int_{0}^{\frac{\pi}{4}} \ln(\cos (u)) \ du \right)=$$
$$\frac{\pi}{2} \ln(2) + 2\left(\int_{0}^{\frac{\pi}{4}} \ln(\sin (u)) \ du+\int_{0}^{\frac{\pi}{4}} \ln(\cos (u)) \ du\right)$$
that may be rewritten as
$$I=\frac{\pi}{2} \ln(2)+2I$$
thus
$$I=-\frac{\pi}{2} \ln(2)$$
Then the final result is $\displaystyle -\frac{\pi^2}{2} \ln(2).$