Let $I = \int_0^\pi x \log \sin xdx$ ……..(i)


$I = \int_0^\pi {(\pi - x)\log \sin (\pi - x)} dx$

$ = \int_0^\pi {(\pi - x)} \log \sin xdx$ …….(ii)

$2I = \pi \int_0^\pi {\log } \sin xdx$ …….(iii)

$2I = 2\pi \int_0^{\pi /2} {\log } \sin xdx$

$I = \pi \int_0^{\pi /2} {\log } \sin xdx$ ………(iv)

Now, $I = \pi \int_0^{\pi /2} {\log } \sin (\pi /2 - x)dx$ …….(v)

On adding Eqs. (iv) and (v), we get $2I = \pi \int_0^{\pi /2} {(\log \sin x + \log \cos x)} dx$

$2I = \pi \int_0^{\pi /2} {\log } \sin x\cos xdx$

$ = \pi \int_0^{\pi /22} {\log } \frac{{2\sin x\cos x}}{2}dx$

$2I = \pi \int_0^{\pi /2} {(\log \sin 2x - \log 2)} dx$

$2I = \pi \int_0^{\pi /2} {\log } \sin 2xdx - \pi \int_0^{\pi /2} {\log } 2dx$

Let's put $2x = t \Rightarrow dx = \frac{1}{2}dt$
As $x \to 0,$ then $t \to 0$

and $x \to \frac{\pi }{2},$ then $t \to \pi $
therefore,$2I = \frac{\pi }{2}\int_0^\pi {\log } \sin tdt - \frac{{{\pi ^2}}}{2}\log 2$

$ \Rightarrow $$2I = \frac{\pi }{2}\int_0^\pi {\log } \sin xdx - \frac{{{\pi ^2}}}{2}\log 2$

$ \Rightarrow $$2I = I - \frac{{{\pi ^2}}}{2}\log 2$ [from eq.(iii)]
therefore,$I = - \frac{{{\pi ^2}}}{2}\log 2 = \frac{{{\pi ^2}}}{2}\log \left( {\frac{1}{2}} \right)$

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