Explanation

(a)
Let $I = \int_{0}^{\pi/2}{\log{\sin{x\ dx}}}$ …(i)
$I = \int_{0}^{\pi/2}{\log{\sin{\left( \frac{\pi}{2} - x \right)dx = \int_{0}^{\pi/2}{\log{\cos{x\ dx}}}}}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi/2}{\log{\sin{x\ \cos{x\ dx}}}}$
$= \int_{0}^{\pi/2}{\log{\sin{2x\ dx - \int_{0}^{\pi/2}{\log{2{\ dx}}}}}}$
$= \frac{1}{2}\int_{0}^{\pi}{\log{\sin{t\ dt - \frac{\pi}{2}\log 2}}}$
$= \frac{1}{2}.2\int_{0}^{\pi/2}{\log{\sin{t\ dt - \frac{\pi}{2}\log 2}}}$
$\Rightarrow 2I = I - \frac{\pi}{2}\log{2\ \Rightarrow I = - \frac{\pi}{2}\log 2}$