Question

Evaluate:
${\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

Answer 1

$I=\int \frac{\pi }{2}(2log\sin x-log\sin 2x)dx$

$I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}[2log\sin x-log(2\sin x\cos x\left)\right]dx$

$I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}[2log\sin x-log\sin x-log\cos x-log2]dx$

$I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}(2log\sin x-2log\sin x-log2)dx$

$I=\frac{{\pi }^2}{2}{\int }_0^{\frac{\pi }{2}}(-log2)dx$

$I=-\frac{{\pi }^2}{4}\left(log2\right)$.