Integrals — Class 12 Maths Solution

.: Let $I = \int {\sin 4x\sin 8x} dx$

$= \cfrac{1}{2}\int {2\sin 4x\sin 8x} dx = \cfrac{1}{2}\int {\left( {\cos 4x - \cos 12x} \right)} dx$

$= \cfrac{1}{2}\left[ {\cfrac{1}{4}\sin 4x - \cfrac{1}{{12}}\sin 12x} \right] + C$