Verifing $\int_0^{\pi}x\ln(\sin x)\,dx=-\ln(2){\pi}^2/2$ 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.
 A: Let $I = \int_0^\pi x \ln(\sin x) \mathrm{d} x$. Then, changing variables $x \to \pi -x$:
$$
    I = \int_0^\pi \left(\pi -x \right) \ln( \sin x) \mathrm{d} x = \pi \int_0^\pi \ln(\sin x) \mathrm{d} x - I
$$
Therefore:
$$
   I = \frac{\pi}{2} \int_0^\pi \ln(\sin x)\mathrm{d} x \stackrel{\text{symmetry}}{=} \pi \int_0^{\pi/2} \ln(\sin x) \mathrm{d} x
$$
The latter integral had been solved elsewhere.
A: 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 $\displaystyle -\frac{\pi^2}{2} \ln(2).$
A: For what is worth, the general case of what Sasha did is
If
$$I=\int_0^\pi xf(\sin x )dx$$
then
$$I=\frac \pi2\int_0^\pi f(\sin x )dx$$
and the argument is completely analogous.
