# Integral of $x \ln( \sin (x))$ from 0 to $\pi$

$$\int_0^\pi x \ln(\sin (x))dx$$

I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$\int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?

By making the change of variable $$u=\pi -x$$ you get that $$I=\int_0^\pi x \ln(\sin x)\:dx=\int_0^\pi (\pi-u) \ln(\sin u)\:du=\pi\int_0^\pi \ln(\sin u)\:du-I$$ giving $$I=\frac{\pi}2\int_0^\pi \ln(\sin u)\:du=\pi\int_0^{\pi/2} \ln(\sin u)\:du.$$ Then conclude with the classic evaluation of the latter integral: see many answers here.
• The pre-factor of the last integral should be $\pi$ instead of $\pi/4$, no? – GDumphart Dec 6 '15 at 16:08