Question

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

Answer 1

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

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

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

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

$={\int }_0^{\pi /2}log\sin x-{\int }_0^{\pi /2}log2dx-{\int }_0^{\pi /2}log\cos (\frac{\pi }{2}-x)dx$

$[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]$

$={\int }_0^{\pi /2}log\sin x-{\int }_0^{\pi /2}log2dx-{\int }_0^{\pi /2}log\sin xdx$

$=-{\int }_0^{\pi /2}log2dx$

$=-log2[x{]}_0^{\pi /2}=-log2[\frac{\pi }{2}-0]$

$\frac{\pi }{2}log\left(\frac{1}{2}\right)$.