Find the integral of the function $\cos 2x \cos 4x \cos 6x$.

We use the trigonometric identity $\cos A \cos B = \frac{1}{2} \{\cos(A+B) + \cos(A-B)\}$.
First,consider the integral $I = \int \cos 2x \cos 4x \cos 6x \, dx$.
Using the identity for $\cos 4x \cos 6x$:
$I = \int \cos 2x \left[ \frac{1}{2} (\cos(4x+6x) + \cos(4x-6x)) \right] dx$
$I = \frac{1}{2} \int \cos 2x (\cos 10x + \cos(-2x)) \, dx$
$I = \frac{1}{2} \int (\cos 2x \cos 10x + \cos^2 2x) \, dx$
Using $\cos^2 2x = \frac{1 + \cos 4x}{2}$ and the identity for $\cos 2x \cos 10x$:
$I = \frac{1}{2} \int \left[ \frac{1}{2} (\cos 12x + \cos(-8x)) + \frac{1 + \cos 4x}{2} \right] dx$
$I = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx$
Integrating term by term:
$I = \frac{1}{4} \left[ \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4} \right] + C$
$I = \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{x}{4} + \frac{\sin 4x}{16} + C$