Question

$\int log\sin xdx$.

Answer 1

$I={\int }_0^{\pi /2}\log \sin xdx$

$={\int }_{\pi /2}^0\log \sin (\frac{\pi }{2}-x)d(\frac{\pi }{2}-x)$

So $I={\int }_0^{\pi /2}\log \cos xdx$

$I+I={\int }_0^{\pi /2}\log \frac{\sin 2x}{2}dx$

$=\frac{1}{2}{\int }_{2x=0}^{\pi }\log \sin 2xd\left(2x\right)-{\int }_0^{\pi /2}\log 2dx$

$=\frac{1}{2}[{\int }_0^{\pi /2}\log \sin xdx+{\int }_{\pi /2}^{\pi }\log \sin xdx]-\frac{\pi }{2}\log 2$

$2I=\frac{1}{2}[I+I]-dfrac\pi 2\log 2$

$I=-\frac{\pi }{2}\log 2$