Explanation

(b)
Let $I = \int_{- \pi}^{\pi}{\frac{2x}{1 + \cos^{2}x}dx + 2\int_{- \pi}^{\pi}{\frac{x\sin x}{1 + \cos^{2}x}dx}}$
$\Rightarrow I = 0 + 4\int_{0}^{\pi}\frac{x\sin{x\ dx}}{1 + \cos^{2}x}$ ...(i)
$\left\lbrack \because\int_{- \pi}^{\pi}{\frac{2x}{1 + \cos^{2}x}dx}\ is\ an\ odd\ function \right\rbrack$
$\Rightarrow I = 4\int_{0}^{\pi}{\frac{(\pi - x)\sin x}{1 + \cos^{2}x}dx}$…(ii)
On adding Eqs. (i) and (ii), we get
$2I = 4\pi\int_{0}^{\pi}{\frac{\sin x}{1 + \cos^{2}x}dx}$
Put $cos\ x = t\ \Rightarrow \ \sin{x\ dx = - dt}$
$\therefore\ I = 2\pi\int_{1}^{- 1}{\frac{- dt}{1 + t^{2}} = 2\pi\lbrack\tan^{- 1}{t\left. \ \ \right\rbrack_{- 1}^{1} = 2\pi\left\lbrack \frac{\pi}{4} + \frac{\pi}{4} \right\rbrack = \pi^{2}}}$