∫limits_0^π / 2 sin 2 x ln tan x d x=

$I=\int\limits_0^{\pi / 2} \sin 2 x \ln (\tan x) d x$            ....(1)

$I=\int\limits_0^{\pi / 2} \sin 2\left(\frac{\pi}{2}-x\right) \ln \tan \left(\frac{\pi}{2}-x\right) d x$

$I =\int\limits_0^{\pi / 2} \sin 2 x \ln (\cot x) d x$            ....(2)

Adding (1) and (2)

$2I=\int\limits_0^{\pi / 2} \sin 2 x[\ln (\tan x)+\ln (\cot x)] d x=\int\limits_0^{\pi / 2} \sin 2 x \ln 1 d x=0$