Question

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

Answer 1

We know that,

$\cos 2x\cos 4x\cos 6x=$ $\cos 2x\cos 4x\cos 6x=\frac{1}{4}(\cos 12x+\cos 8x+\cos 4x+1)$

Taking integration both side ,

$\int \cos 2x\cos 4x\cos 6x=\int \frac{1}{4}(\cos 12x+\cos 8x+\cos 4x+1)dx$

$=\frac{1}{4}[\int \cos 12xdx+\int \cos 8xdx+\int \cos 4xdx+\int 1dx]$

$=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+C$