Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(2\log \sin x-\log \sin 2x\right)dx$

Answer 1

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

we have $f\left(x\right)=2\log \sin x-\log \sin 2x$

$f\left(x\right)=\log \frac{{\sin }^{2}x}{\sin 2x}$

$=\log \tan x-\log 2$

$I_1={\int }_0^{\frac{\pi }{2}}\log \tan xdx$

Let $y=\frac{\pi }{2}-x,dy=-dx,\tan x=\cot y$

$I_1=-{\int }_0^{\frac{\pi }{2}}\log \cot ydy=-{\int }_0^{\frac{\pi }{2}}\log \tan xdx=-I$

$\Rightarrow 2I_1=0$

$\Rightarrow I_1=0$

Now $I={\int }_0^{\frac{\pi }{2}}\log \tan xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$

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