Question

Evaluate :
$\int \cos 2x\cos 4x\cos 6xdx$

Answer 1

$I=\int cos2xcos4xcos6xdxcos2xcos4xcos6x=\frac{1}{2}\left(\right(2cos2xcos4x)cos6x)=\frac{1}{2}\{(cos6x+cos2x)cos6x\}=\frac{1}{2}\{\frac{1}{2}\left(2{\cos }^{2}6x\right)+\frac{1}{2}\left(2\cos 2x\cos 6x\right)\}=\frac{1}{4}(1+\cos 12x+\cos 8x+\cos 4x)I=\frac{1}{4}\int (1+\cos 4x+\cos 8x+\cos 12x)dx=\frac{1}{4}\{x+\frac{sin4x}{4}+\frac{sin8x}{8}+\frac{sin12x}{12}\}+C$