Question

Integrate $\int \cos 2x\cos 4x\cos 6xdx$.

Answer 1

Solve the given integral

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

We know that,

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

So for$\int \cos 2x\cos 4x\cos 6xdx$, we can write as,

$=\int \cos 2x.\frac{1}{2}\left[\cos 10x+\cos 2x\right]dx$

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

$$\begin{gathered} \mathbf{}\mathbf{\because }\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{1} \\ \mathbf{\Rightarrow }{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{1}}{\mathbf{2}} \end{gathered}$$

So, we have

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

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

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

Hence, Integration of $\int \cos 2x\cos 4x\cos 6xdx$ is $\frac{1}{4}\left[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x\right]+C$, where $C$ is the arbitrary constant.