It should be in the sub-forum "Interesting Math Problem" but there is no such place for me to post (Rofl) It is not very hard but the answer is extremely elegant !!

Show that $\displaystyle I_{m} = \int_0^{\frac{\pi}{2}} \ln[\sin(x)] \sin^{2m}(x)~dx = \frac{\pi \binom{2m}{m} }{4^m} [( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - ... + \frac{1}{2m-1} - \frac{1}{2m} )- \ln2 ] $

and $\displaystyle I_0 = -\frac{\pi}{2}\ln2 $