Question

${\int }_{\cos 2x\cos 4x\cos 6xdx}$

Answer 1

It is known that, $\cos A\cos B=\frac{1}{2}\left\{\cos \right(A+B)+\cos (A-B\left)\right\}$
$\therefore \int \cos 2x\left(\cos 4x\cos 6x\right)dx=\int \cos 2x\left[\frac{1}{2}\left\{\cos \right(4x+6x)+\cos (4x-6x\left)\right\}\right]dx$
$=\frac{1}{2}\int \{\cos 2x\cos 10x+\cos 2x\cos (-2x\left)\right\}dx$
$=\frac{1}{2}\int \{\cos 2x\cos 10x+{\cos }^{2}2x\}dx$
$=\frac{1}{2}\int [\left\{\frac{1}{2}\cos \right(2x+10x)+\cos (2x-10x\left)\right\}+\left(\frac{1+\cos 4x}{2}\right)]dx$
$=\frac{1}{4}\int (\cos 12x+\cos 8x+1+\cos 4x)dx$
$=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+x+\frac{\sin 4x}{4}]+C$