Question

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

Answer 1

\[\int\sin x .\sin 2x .\text{ sin  3x  dx }\]
\[ = \frac{1}{2}\int\left( 2 \sin 2x \cdot \sin x \right) \text{ sin  3x  dx }\]
\[ = \frac{1}{2}\int\left[ \text{ cos} \left( 2x - x \right) - \text{ cos } \left( 2x + x \right) \right] \text{ sin  3x  dx }.............. \left[ \because \text{ 2  sin  A  sin  B = cos (A - B) - cos (A + B)} \right]\]
\[ \Rightarrow  = \frac{1}{2}\int\left[ \cos x - \cos 3x \right] \text{ sin  3x  dx }\]
\[ = \frac{1}{2}\int\sin 3x \cdot \text{ cos  x  dx } - \frac{1}{2}\int\sin 3x \cdot \text{ cos  3x dx }\]
\[ = \frac{1}{4}\int \text{ 2 }\sin 3x \cdot \text{ cos  x   dx} - \frac{1}{4}\int\text{ 2 }\sin 3x \cdot \text{ cos  3x  dx } \]
\[ = \frac{1}{4}\int\left[ \sin 4x + \sin 2x \right]dx - \frac{1}{4}\int\text{ sin  6x  dx } ............. \left[ \because \text{ 2  sin  A cos  B = sin (A + B) - sin (A - B)} \right]\]
\[ = \frac{1}{4}\left[ \frac{- \cos 4x}{4} - \frac{\cos 2x}{2} \right] - \frac{1}{4}\left[ - \frac{\cos 6x}{6} \right] + C\]
\[ = - \frac{\cos 4x}{16} - \frac{\cos 2x}{8} + \frac{\cos 6x}{24} + C\]