Question

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

Answer 1

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

$f\left(x\right)=\log \left[{\sin }^{2}x/\right(2\sin x\cos x\left)\right]$

$f\left(x\right)=\log \tan x=\log 2$

$T_1={\int }_{x=0}^{\pi /2}\log \tan x$

$y=\pi /2-x,dy=-dx$

$\therefore tanx=\cot y$

$I_1={\int }_{y=0}^{\pi /2}\log \cot ydy$

$=-{\int }_{y=0}^{\pi /2}\log \tan ydy$

$=-I_1$

$\therefore I_1=0$

$\therefore I={\int }_0^{\pi /2}-\log 2dx=-\pi /2\log 2$