Question

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

Answer 1

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]
`= 1/2 ∫ (2   sin  2x   sin 4x )   sin 6x  dx `
\[ =  \frac{1}{2}\int\left[ \text{cos}\left( 2x - 4x \right) - \text{cos}\left( 2x + 4x \right) \right] \text{sin 6x dx}\]
\[ = \frac{1}{2}\int\left[ \text{cos}\left( 2x \right) - \text{cos}\left( 6x \right) \right] \text{sin 6x dx}\]
\[ = \frac{1}{2}\left[ \int\text{cos}\left( 2x \right)\text{sin}\left( 6x \right) dx - \int\text{cos}\left( 6x \right)\text{sin}\left( 6x \right) dx \right]\]
\[ = \frac{1}{4}\left[ \int2\text{cos}\left( 2x \right)\text{sin}\left( 6x \right) dx - \int2\text{cos}\left( 6x \right)\text{sin}\left( 6x \right) dx \right]\]
\[ = \frac{1}{4}\left\{ \int\left[ \text{sin}\left( 2x + 6x \right) - \text{sin}\left( 2x - 6x \right) \right] dx - \int\text{sin}\left( 12x \right) dx \right\}\]
\[ = \frac{1}{4}\left[ \int\text{sin}\left( 8x \right) dx + \int\text{sin}\left( 4x \right) dx - \int\text{sin}\left( 12x \right) dx \right]\]
\[ = \frac{1}{4}\left[ \frac{- \text{cos}\left( 8x \right)}{8} + \frac{- \text{cos}\left( 4x \right)}{4} + \frac{\text{cos}\left( 12x \right)}{12} \right] + c\]
\[ = - \frac{\text{cos}\left( 8x \right)}{32} - \frac{\text{cos}\left( 4x \right)}{16} + \frac{\text{cos}\left( 12x \right)}{48} + c\]

Hence, \[\int\text{sin  2x   sin 4x    sin 6x   dx }= - \frac{\cos\left( 8x \right)}{32} - \frac{\cos\left( 4x \right)}{16} + \frac{\cos\left( 12x \right)}{48} + c\]