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

$\frac{\pi}{3}$

The correct answer is Option (1) → $\frac{\pi}{3}$

Let $I = \int\limits_{0}^{\pi} x \sin x \cos^2 x \, dx  \dots(i)$

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

$\Rightarrow I = \int\limits_{0}^{\pi} (\pi - x) \sin x \cos^2 x \, dx  \dots(ii)$

On adding Eqs. (i) and (ii), we get

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

Put $\cos x = t$

$\Rightarrow -\sin x \, dx = dt$

As $x \to 0, \text{ then } t \to 1$

and $x \to \pi, \text{ then } t \to -1$

$∴2I = -\pi \int\limits_{1}^{-1} t^2 \, dt \Rightarrow 2I = -\pi \left[ \frac{t^3}{3} \right]_1^{-1}$

$\Rightarrow 2I = -\frac{\pi}{3} [-1 - 1] \Rightarrow 2I = \frac{2\pi}{3}$

$∴I = \frac{\pi}{3}$