TITLE: Integral of $ x \ln( \sin (x))$ from 0 to $ \pi $
QUESTION [6 upvotes]: $$\int_0^\pi x \ln(\sin (x))dx  $$

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?

REPLY [12 votes]: 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.