Question

\[\int\limits_0^\pi x \sin^3 x\ dx\]

Answer 1

\[Let\ I = \int_0^\pi x \sin^3 x\ d x . . . (i)\]
\[ = \int_0^\pi \left( \pi - x \right) \sin^3 \left( \pi - x \right) d x\]
\[ = \int_0^\pi \left( \pi - x \right) \sin^3 x dx . . . (ii)\]
\[\text{Adding (i) and (ii) we get}\]
\[2I = \int_0^\pi \left( x + \pi - x \right) \sin^3 x\ d x\]
\[ = \int_0^\pi \pi \sin^3 x d x\]
\[ = \int_0^\pi \pi \frac{3 \sin x - \sin 3x}{4} d\ x\]
\[ = \frac{\pi}{4} \int_0^\pi \left( 3 \sin x - \sin 3x \right) d x\]
\[ = \frac{\pi}{4} \left[ - 3 \cos x + \frac{\cos 3x}{3} \right]_0^\pi \]
\[ = \frac{\pi}{4}\left[ - 3 \cos \pi + 3\cos 0 + \frac{\cos 3\pi}{3} - \frac{\cos 0}{3} \right]\]
\[ = \frac{\pi}{4}\left[ 3 + 3 + \frac{- 1}{3} - \frac{1}{3} \right]\]
\[ = \frac{\pi}{2}\left[ 3 - \frac{1}{3} \right]\]
\[ = \frac{\pi}{2} \times \frac{8}{3}\]
\[ = \frac{4\pi}{3}\]
\[ \therefore I = \frac{2\pi}{3}\]