∫^(π)/2_0sin2x log tan x dx =

The correct answer is Option (3) → $0$

$I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x \log \tan x \, dx$  ...(1)

$I=\int\limits^{\frac{\pi}{2}}_{0}(π-2x)\log\tan(\frac{π}{2}-x)dx$

$I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log\cot xdx$   ...(2)

eq. (1) + eq. (2)

$2I=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log\tan x\cot xdx$

$=\int\limits^{\frac{\pi}{2}}_{0}\sin 2x\log 1dx$

$=\int\limits^{\frac{\pi}{2}}_{0}0dx=0$

$I=0$