INTEGRALS — Maths STD 12 Science — Question

$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$

$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$

$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$

$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$

$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$

$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$

$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$

$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$

$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$

$=\frac{\pi}{2}\times\frac{8}{3}$

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

$\therefore\ \text{I}=\frac{2\pi}{3}$