Prove

It is known that, $\cos A \cos B=\frac{1}{2}\{\cos (A+B)+\cos (A-B)\}$

$\therefore \int \cos 2 x(\cos 4 x \cos 6 x) d x=\int \cos 2 x\left[\frac{1}{2}\{\cos (4 x+6 x)+\cos (4 x-6 x)\}\right] d x$

$=\frac{1}{2} \int\{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)\} d x$

$=\frac{1}{2} \int\left\{\cos 2 x \cos 10 x+\cos ^{2} 2 x\right\} d x$

$=\frac{1}{2} \int\left[\left\{\frac{1}{2} \cos (2 x+10 x)+\cos (2 x-10 x)\right\}+\left(\frac{1+\cos 4 x}{2}\right)\right] d x$

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

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