Answer:
Option A
Explanation:
Let $I=\int_{0}^{\pi/2} (\log \tan x).\sin 2 x dx$ ......(i)
$I=\int_{0}^{\pi/2} \log \tan(\frac{\pi}{2}- x).\sin 2(\frac{\pi}{2}- x) dx$
$\left[\because\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx\right]$
$\Rightarrow I=\int_{0}^{\pi/2} \log \cot x.\sin 2 x dx$...........(ii)
$[\because\sin (\pi-2x)=\sin 2x]$
On adding eqs(i) and (ii) , we get
$2I\int_{0}^{\pi/2} \log\tan x.\sin2x dx+\int_{0}^{\pi/2} \log\cot x\sin 2x dx$
= $\int_{0}^{\pi/2} \sin2x\log(\tan x.\cot x) dx$
$[\because$ $ \log m+\log n=\log (m.n)]$
$\int_{0}^{\pi/2} \sin 2x\log1 dx$
$\Rightarrow$ $1=0[\because log 1=0]$
$\therefore $ $\int_{0}^{\pi/2} \sin 2x\log (\tan x) dx=0$
