Explanation

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