Question

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

Answer 1

Consider the given integral.

$I=\int \cos 2x\cos 4xdx$

We know that

$\cos A\cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)]$

Therefore,

$I=\frac{1}{2}\int [\cos (2x+4x)+\cos (2x-4x)]dx$

$I=\frac{1}{2}\int [\cos 6x+\cos (-2x)]dx$

$I=\frac{1}{2}\int [\cos 6x+\cos 2x]dx$

$I=\frac{1}{2}[\frac{\sin 6x}{6}+\frac{\sin 2x}{2}]+C$

$I=\frac{\sin 6x}{12}+\frac{\sin 2x}{4}+C$

Hence, this is the answer.