Question

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

Answer 1

$I={\int }_0^{\pi /2}\sin 2x\log \tan xdx$
Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$\therefore I={\int }_0^{\pi /2}\sin 2(\pi /2-x)\log \tan (\pi /2-x)dx={\int }_0^{\pi /2}\sin 2x\log \cot xdx=-{\int }_0^{\pi /2}\sin 2x\log \tan xdx\therefore I=-I\Rightarrow 2I=0\Rightarrow I=0$