Explanation

(c)
Let $\int_{0}^{\pi/2}{\sin{2x\log{\tan{x\ dx}}}}$ …(i)
$I = \int_{0}^{\pi/2}{\sin{2\left( \frac{\pi}{2} - x \right)\log{\tan{\left( \frac{\pi}{2} - x \right)dx}}}}\ \ \ \left\lbrack \because\int_{0}^{a}{f(x)dx}\int_{0}^{a}{f(a - x)dx} \right\rbrack$
$I = \int_{0}^{\pi/2}{\sin{2x}}\log{\cot{x\ dx}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi/2}{\sin{2x}}\log{(\tan{x\cot{x)dx}}}$
$= \int_{0}^{\pi/2}{\sin{2x}}\log{1\ dx = \int_{0}^{\pi/2}{0\ dx}}$
$\Rightarrow I = 0$