Question

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

Answer 1

$Given, $\int \cos 2x\cos 4x\cos 6xdx$

We know that,

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

And $\cos 2\theta =2{\cos }^2\theta -1\Rightarrow {\cos }^2\theta =\frac{1}{2}(\cos 2\theta -1)$

$\therefore \int \cos 2x\cos 4x\cos 6xdx=\int \frac{1}{2}[\cos (2x+4x)+\cos (2x-4x)]\cos 6xdx=\frac{1}{2}\int [\cos 6x+\cos (-2x)]\cos 6xdx=\frac{1}{2}\int [\cos 6x+\cos 2x]\cos 6xdx=\frac{1}{2}\int [{\cos }^{2}6x+\cos 2x\cos 6x]dx=\frac{1}{2}\int \frac{1}{2}\left[\right(\cos 12x+1)+\cos 8x+\cos 4x]dx=\frac{1}{4}\int [\cos 12x+\cos 8x+\cos 4x+1]dx=\frac{1}{4}[\int \cos 12xdx+\int \cos 8xdx+\int \cos 4xdx+\int \left(1\right)dx]=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+C\therefore \int \cos 2x\cos 4x\cos 6xdx=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+C.$