A) \[\pi \]
B) \[\pi /2\]
C) 0
D) \[2\pi \]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /2}{\,\,\,\,\,\sin 2x\log \tan x\,\,dx}\], \[I=\int_{0}^{\pi /2}{\sin 2\,\left( \frac{\pi }{2}-x \right)\log \tan \left( \frac{\pi }{2}-x \right)\,\,dx}\], \[[\because \int_{0}^{a}{f\,(x)\,dx=\int_{0}^{a}{f\,(a-x)\,dx]}}\] \[=\int_{0}^{\pi /2}{\,\,\,\,\,\sin 2x\log \cot x\,\,dx}\]\[=-\int_{0}^{\pi /2}{\,\,\,\,\,\sin 2x\log \tan x\,\,dx}\] \[\therefore I=-I\,\,\]Þ 2I = 0 \[\Rightarrow I=0.\]
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