Evaluate $\int\limits_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$

$\frac{\pi^2}{4}$

The correct answer is Option (2) → $\frac{\pi^2}{4}$

Let $I = \int\limits_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$. Then, by $P_4$, we have

$I = \int\limits_{0}^{\pi} \frac{(\pi - x) \sin (\pi - x) \, dx}{1 + \cos^2 (\pi - x)}$

$= \int\limits_{0}^{\pi} \frac{(\pi - x) \sin x \, dx}{1 + \cos^2 x} = \pi \int\limits_{0}^{\pi} \frac{\sin x \, dx}{1 + \cos^2 x} - I$

or $2I = \pi \int\limits_{0}^{\pi} \frac{\sin x \, dx}{1 + \cos^2 x}$

or $I = \frac{\pi}{2} \int\limits_{0}^{\pi} \frac{\sin x \, dx}{1 + \cos^2 x}$

Put $\cos x = t$ so that $-\sin x \, dx = dt$. When $x = 0, t = 1$ and when $x = \pi, t = -1$.

Therefore, (by $P_1$) we get

$I = \frac{-\pi}{2} \int\limits_{1}^{-1} \frac{dt}{1 + t^2} = \frac{\pi}{2} \int\limits_{-1}^{1} \frac{dt}{1 + t^2}$

$= \pi \int\limits_{0}^{1} \frac{dt}{1 + t^2} \text{ (by } P_7, \text{ since } \frac{1}{1 + t^2} \text{ is even function)}$

$= \pi \left[ \tan^{-1} t \right]_{0}^{1} = \pi [\tan^{-1} 1 - \tan^{-1} 0] = \pi \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi^2}{4}$