A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) 1
D) \[2\pi \]
E) 0
Correct Answer: E
Solution :
Let \[I=\int_{0}^{\pi /2}{\sin 2x\log \tan x\,dx}\] ...(i) \[\therefore \]\[I=\int_{0}^{\pi /2}{\sin (\pi -2x)\log \tan \left( \frac{\pi }{2}-x \right)}dx\] \[I=\int_{0}^{\pi /2}{\sin 2x\log \cot x\,\,}dx\] ...(ii) From Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\sin 2x\log \tan x\,\cot x\,}dx\] \[=\int_{0}^{\pi /2}{\sin 2x\log 1}dx\] \[=0\] \[\Rightarrow \] \[I=0\]
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