Find the value of the integral is `int_(0)^(pi) x log sinx dx`

Explanation

Let `I=int_(0)^(pi)x log sin x dx`………………….1
`:.I=int_(0)^(pi)(pi-x)logsin(pi-x)dx`
or `I=int_(0)^(pi)(pi-x)logsinxdx`……………..2
Adding 1 and 2 we get
`2pi=piint_(0)^(pi)logsinxdx`
or `2I=2pi int_(0)^(pi//2)log sin x dx`
or `I=pi int_(0)^(pi//2)log sinx`
`=pi{1/2 pi log (1//2)}=1/2pi^(2)log(1//2)`