Explanation

(c)
Let $I = \int_{0}^{\pi/2}{\log{\sin{x\ dx\ldots(i)}}}$
$I = \int_{0}^{\pi/2}{\log{\sin{\left( \frac{\pi}{2} - x \right)dx}}}$
$= \int_{0}^{\pi/2}{\log{\cos x}}\ dx$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi/2}{(\log{\sin{x + \log{\cos{x)dx}}}}}$
$= \int_{0}^{\pi/2}{\log{{sin2}{x + {dx -}{\log{2\int_{0}^{\pi/2}{dx}}}}}}$
Put $2x = t\ \Rightarrow \ dx = \frac{dt}{2}$
$\therefore 2I = \int_{0}^{\pi}\frac{\log{\sin t}}{2}dt - \log{2\lbrack x\rbrack_{0}^{\pi/2}}$
$= \frac{1}{2} \times 2\int_{0}^{\pi/2}{\log{\sin{t\ dt - \frac{\pi}{2}\log 2}}}$
$= \int_{0}^{\pi/2}{\log{\sin{xdx - \frac{\pi}{2}\log 2}}}$
$= I - \frac{\pi}{2}\log 2$
$\Rightarrow \ I = - \frac{\pi}{2}\log 2$