Explanation

(d)
Let $I = \int_{0}^{\pi/2}{\frac{dx}{1 + \cot x} = \int_{0}^{\pi/2}{\frac{\cos x}{\cos{x + \sin x}}dx}}$ …(i)
$\Rightarrow I = \int_{0}^{\pi/2}\frac{\cos\left( \frac{\pi}{2} - x \right)}{\cos{\left( \frac{\pi}{2} - x \right) + \sin\left( \frac{\pi}{2} - x \right)}}dx$
$\Rightarrow I = \int_{0}^{\pi/2}\frac{\sin x}{\cos{x + \sin x}}dx$ ...(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi/2}{1\ dx = \frac{\pi}{2}}$
$\Rightarrow \ I = \frac{\pi}{4}$