Explanation

(d)
Let $I = \int_{0}^{\pi}{\frac{\theta\sin\theta}{1 + \cos^{2}\theta}d\theta}$ …(i)
$\Rightarrow I = \int_{0}^{\pi}{\frac{(\pi - \theta)\sin\theta}{1 + \cos^{2}\theta}d\theta\ \ldots(ii)}$
On adding Eqs. (i) and (ii), we get
$2I = \int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^{2}\theta}\ d\theta$
Put $\cos{\theta = t \Rightarrow - \sin{\theta d\theta = dt}}$
$\therefore 2I = - \pi\int_{1}^{- 1}{\frac{1}{1 + t^{2}}dt = 2\pi\int_{0}^{1}{\frac{1}{1 + t^{2}}dt}}$
$= 2\pi\lbrack\tan^{- 1}{t\left. \ \ \right\rbrack_{0}^{1} = 2\pi.\frac{\pi}{4}}$
$\Rightarrow \ I = \frac{\pi^{2}}{4}$