MCQ
$\int_{\,0}^{\,\pi /2} {\sin 2x\log \tan x\,dx} $ is equal to
- A$\pi $
- B$\pi /2$
- ✓$0$
- D$2\pi $
$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} $
$\therefore I = - I\,\,==> 2I = 0$ $ \Rightarrow I = 0.$
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