The value of the integral $\int\limits^{\pi}_{0}2x\, sin^3x\, dx $ is :

$\frac{4\pi}{3}$

The correct answer is Option (3) → $\frac{4\pi}{3}$

$I=\int\limits^{\pi}_{0}2x\sin^3x\, dx$   ...(1)

$I=\int\limits^{\pi}_{0}2(π-x)\sin^3(π-x)dx=\int\limits^{\pi}_{0}2(π-x)\sin^3xdx$  ...(2)

Adding (1) and (2)

$2I=2π\int\limits^{\pi}_{0}\sin^3xdx$

$I=π\int\limits^{\pi}_{0}\sin^3xdx$

as $\sin^3x=3\sin x-4\sin^3x$

$\sin^3x=\frac{3\sin x-\sin^3x}{4}$

$I=π\int\limits^{\pi}_{0}\frac{3\sin x}{4}-\frac{\sin^3x}{4}dx$

$=π\left[\frac{\cos^3x}{12}-\frac{3\cos x}{4}\right]^{\pi}_{0}$

$=π\left[-\frac{1}{12}-\frac{1}{12}+\frac{3}{4}+\frac{3}{4}\right]$

$=π\left[-\frac{1}{6}+\frac{3}{2}\right]=\frac{4\pi}{3}$