${\int }_0^{\pi /2} \sin 2x\log \tan xdx$ is equal to 

Solution:

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

Then,

$\begin{array}{l}I{\int }_0^{\pi /2} \sin 2\left(\frac{\pi }{2}-x\right)\log \tan \left(\frac{\pi }{2}-x\right)dx\\ \Rightarrow I={\int }_0^{\pi /2} \sin 2x\log \cot xdx\end{array}$

Adding (i) and (ii), we get

$\begin{array}{l}2I={\int }_0^{\pi /2} \sin 2x(\log \tan x+\log \cot x)dx\\ \Rightarrow 2I={\int }_0^{\pi /2} \sin 2x\times \log 1dx\Rightarrow 2I=0\Rightarrow I=0\end{array}$