Question

Integrate the following integrals:

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]

Answer 1

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]
` = 1/2  ∫   ( 2  sin x  cos  2x)  sin 3x  dx` 
\[ = \frac{1}{2}\int\left[ \text{sin}\left( x + 2x \right) + \text{sin}\left( x - 2x \right) \right] \text{sin}\left( 3x \right) dx\]
\[ = \frac{1}{2}\int\left[ \text{sin}\left( 3x \right) - \text{sin}\left( x \right) \right] \text{sin}\left( 3x \right) dx\]
\[ = \frac{1}{2}\left[ \int \text{sin}^2 \left( 3x \right) dx - \int\text{sin}\left( x \right)\text{sin}\left( 3x \right) dx \right]\]
\[ = \frac{1}{4}\left[ \int2 \text{sin}^2 \left( 3x \right) dx - \int2\text{sin}\left( x \right)\text{sin}\left( 3x \right) dx \right]\]
\[ = \frac{1}{4}\left\{ \int\left[ 1 - \text{cos}\left( 6x \right) \right] dx - \int\left[ \text{cos}\left( x - 3x \right) - \text{cos}\left( x + 3x \right) \right] dx \right\}\]
\[ = \frac{1}{4}\left[ \int1 dx - \int\text{cos}\left( 6x \right) dx - \int\text{cos}\left( 2x \right) dx + \int\text{cos}\left( 4x \right) dx \right]\]
\[ = \frac{1}{4}\left[ x - \frac{\text{sin}\left( 6x \right)}{6} - \frac{\text{sin}\left( 2x \right)}{2} + \frac{\text{sin}\left( 4x \right)}{4} \right] + c\]
\[ = \frac{x}{4} - \frac{\text{sin}\left( 6x \right)}{24} - \frac{\text{sin}\left( 2x \right)}{8} + \frac{\text{sin}\left( 4x \right)}{16} + c\]

Hence, 

\[\int\text{ sin}\text{ x }\text{cos 2x   sin 3x}\ dx = \frac{x}{4} - \frac{\text{sin}\left( 6x \right)}{24} - \frac{\text{sin}\left( 2x \right)}{8} + \frac{\text{sin}\left( 4x \right)}{16} + c\]