Question

\[\int x \sin x \cos 2x\ dx\]

Answer 1

\[\int x . \cos 2x \text{ sin x dx }\]
\[ = \frac{1}{2} \int x \left( 2 \cos 2x \sin x \right) dx \left[ \because 2 \cos A \sin B = \sin \left( A + B \right) - \sin \left( A - B \right) \right]\]
\[ = \frac{1}{2} \int x \left( \sin 3x - \sin x \right) dx\]
\[ = \frac{1}{2} \int x . \text{ sin 3x dx }- \frac{1}{2} \int x \text{ sin x dx}\]
\[ = \frac{1}{2}\left[ x\int\text{ sin 3x dx }- \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin 3x dx} \right\}dx \right] - \frac{1}{2}\left[ x\int\text{ sin x dx }- \int\left\{ \frac{d}{dx}\left( x \right)\int\text{ sin x dx }\right\}dx \right]\]
\[ = \frac{1}{2}\left[ x . \left( \frac{- \cos 3x}{3} \right) - \int 1 . \left( \frac{- \cos 3x}{3} \right)dx \right] - \frac{1}{2}\left[ x . \left( - \text{ cos x} \right) - \int 1 . \left( - \cos x \right) dx \right]\]
\[ = \frac{1}{2}\left[ x . \left( \frac{- \cos 3x}{3} \right) + \frac{1}{9}\sin 3x \right] - \frac{1}{2}\left[ x . \left( - \cos x \right) + \sin x \right]\]
\[ = - \frac{x \cos 3x}{6} + \frac{\sin 3x}{18} + \frac{x \cos x}{2} - \frac{\sin x}{2} + C\]