# Find the integrals of the functions$\cos^42x$

Toolbox:
• $(i)\sin^2x=\frac{(1-\cos 2x)}{2}.$
• $(ii)\int\cos ^2xdx=\frac{(1+\cos2 x)}{2}+c.$
Given:$I=\int\cos^42xdx.$

But $\cos^22x=\frac{1+\cos 4x}{2}.$

This can be written as:

$I=\int(\cos^22x)^2dx=\int\bigg(\frac{1+\cos 4x}{2}\bigg)^2dx.$

Taking $\frac{1}{4}$ as the common factor

$\;\;\;=\frac{1}{4}\int(1+\cos 4x)^2dx.$

On separating the terms

$\;\;\;=\frac{1}{4}\int(1+\cos^24x-2\cos4x)dx.$

But $\cos^24x=\frac{(1+\cos 8x)}{2}$.

$\;\;\;=\frac{1}{4}\int[1+\bigg(\frac{1+\cos 8x}{2}\bigg)-2\cos 4x]dx.$

On separating the terms we get,

$I=\frac{1}{4}\int[\frac{3}{2}+\frac{1}{2}\cos 8x+\frac{1}{2}\cos 4x]dx.$

On integrating we get,

$I=\frac{3}{8}x+\frac{1}{8}\frac{1}{8}\sin 8x+\frac{1}{8}\sin 4x+c.$

On multiplying the factors

$\;\;\;=\frac{3}{8}x+\frac{\sin 8x}{64}+\frac{1}{8}\sin 4x+c.$