Explanation

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