Question

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

Answer 1

To find: $\int \cos 2x\cos 4x\cos 6xdx$

Multiply and divide by $2$

$=\frac{2}{2}\int \left(\cos 2x\cos 4x\cos 6x\right)dx$

$=\frac{1}{2}\int \left[\right(2\cos 2x\cos 4x\left)\cos 6x\right]dx$

$=\frac{1}{2}\int (\cos 6x+\cos 2x)\cos 6xdx$

$[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\cos \right(-x)=\cos x]$

$=\frac{1}{2}\int \left[\right(\cos 6x{)}^2+(\cos 2x.\cos 6x)]dx$

Multiply and divide by $2$

$=\frac{1}{4}\int [2{\cos }^{2}6x+2(\cos 2x.\cos 6x\left)\right]dx$

$=\frac{1}{4}\int [1+\cos 12x+(\cos 8x+\cos 4x\left)\right]dx$

$[\because 1+\cos 2\theta =2{\cos }^2\theta ]$

$=\frac{1}{4}[\int 1dx+\int \cos 12xdx+\int \cos 8xdx+\int \cos 4xdx]$

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

Where $C$ is constant of integration.