Explanation

(d)
Let $I = \int_{0}^{\pi}{\lbrack\cot{x\rbrack dx}\ \ldots(i)}$
$\Rightarrow I = \int_{0}^{\pi}{\left\lbrack \cot(\pi - x) \right\rbrack dx = \int_{0}^{\pi}{\lbrack - \cot{x\rbrack dx\ \ \ \ \ldots(ii)}}}$
On adding Eqs. (i) and (ii),
$2I = \int_{0}^{\pi}{\lbrack\cot{x\rbrack dx + \int_{0}^{\pi}{\lbrack - \cot{x\rbrack dx}}}}$
$= \int_{0}^{\pi}{( - 1)dx}$
$\lbrack\because\lbrack x\rbrack + \lbrack - x\rbrack = - ,$if$x \notin z\ an$d$\lbrack x\rbrack + \lbrack - x\rbrack = 0,$if$x \in z\rbrack$
$= \lbrack - x\rbrack_{0}^{\pi} = - \pi$
$\therefore\ I = - \frac{\pi}{2}$