Question

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

Answer 1

\[\int x \cdot \sin^3 x\ dx \]
\[ = \int x \cdot \left[ \frac{1}{4}\left( 3 \sin x - \sin 3x \right) \right] dx \left[ \sin^3 A - \frac{1}{4}\left\{ 3 \sin A - \sin \left( 3A \right) \right\} \right]\]
\[ = \frac{3}{4}\int x_I \cdot \sin_{II} \text{ x dx} - \frac{1}{4}\int x \cdot \text{ sin  3x   dx}\]
\[ = \frac{3}{4}\left[ x\int\text{ sin  x  dx} - \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin  x  dx } \right\}dx \right] - \frac{1}{4}\left[ x\int\text{ sin  3x  dx} - \int\left\{ \frac{d}{dx}\left( x \right)\int\sin 3x dx \right\}dx \right]\]
\[ = \frac{3}{4}\left[ x \left( - \cos x \right) - \int1 \cdot \left( - \cos x \right)dx \right] - \frac{1}{4}\left[ x \left( \frac{- \cos 3x}{3} \right) - \int1 \cdot \left( \frac{- \cos 3x}{3} \right)dx \right]\]
\[ = \frac{- 3x}{4} \cos x + \frac{3}{4} \sin x + \frac{x \cos 3x}{12} - \frac{\sin 3x}{36} + C\]
\[ = \frac{1}{4}\left[ - 3x \cos x + 3\sin x + \frac{x \cos 3x}{3} - \frac{\sin 3x}{9} + C \right]\]