Integrals — Class 12 Maths Solution

: Let$I = \int {\cos 2x\cos 4x\cos 6x\,dx}$

$= \cfrac{1}{2}\int {\left( {2\cos 2x\cos 4x} \right)\cos 6x} \,dx = \cfrac{1}{2}\int {\left( {\cos 6x + \cos 2x} \right)\cos 6x} \,dx$

$= \cfrac{1}{4}\int {2{{\cos }^2}6xdx} + \cfrac{1}{4}\int {\left( {2\cos 2xcos6x} \right)dx}$

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

$= \cfrac{1}{4}x + \cfrac{1}{4}\left( {\cfrac{{\sin 12x}}{{12}}} \right) + \cfrac{1}{4}\left( {\cfrac{{\sin 8x}}{8} + \cfrac{{\sin 4x}}{4}} \right) + C$

$= \cfrac{1}{4}\left[ {x + \cfrac{1}{{12}}\sin 12x + \cfrac{1}{8}\sin 8x + \cfrac{1}{4}\sin 4x} \right] + C$