Explanation

(d)
Let $I = \int_{0}^{\pi/2}{\frac{1}{1 + \tan^{3}x}dx = \int_{0}^{\pi/2}{\frac{1}{1 + \frac{\sin^{3}x}{\cos^{3}x}}dx}}$
$= \int_{0}^{\pi/2}{\frac{\cos^{3}x}{\cos^{3}{x + \sin^{3}x}}dx}$ …(i)
$= \int_{0}^{\pi/2}{\frac{\cos^{3}\left( \frac{\pi}{2} - x \right)}{\cos^{3}{\left( \frac{\pi}{2} - x \right) + \sin^{3}\left( \frac{\pi}{2} - x \right)}}dx}$
$\Rightarrow I = \int_{0}^{\pi/2}{\frac{\sin^{3}x}{\sin^{3}{x + \cos^{3}x}}dx}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{0}^{\pi/3}{\frac{\sin^{3}{x + \cos^{3}x}}{\sin^{3}{x + \cos^{3}x}}dx}$
$\Rightarrow \ \ 2I = \int_{0}^{\pi/2}{1\ dx}$
$\Rightarrow 2I = \lbrack x\rbrack_{0}^{\pi/2} = \frac{\pi}{2}\ \Rightarrow I = \frac{\pi}{4}$