Question

Find the integrals of the functions ${\cos }^{4}2x$

Answer 1

${\cos }^{4}2x=({\cos }^{2}2x{)}^2$
$={\left(\frac{1+\cos 4x}{2}\right)}^2$
$=\frac{1}{4}[1+{\cos }^{2}4x+2\cos 4x]$
$=\frac{1}{4}[1+\left(\frac{1+\cos 8x}{2}\right)+2\cos 4x]$
$=\frac{1}{4}[1+\frac{1}{2}+\frac{\cos 8x}{2}+2\cos 4x]$
$\int co{s}^{4}2x=\frac{1}{4}\int [\frac{3}{2}+\frac{\cos 8x}{8}+\frac{\cos 4x}{2})dx$
$=\frac{3}{8}x+\frac{\sin 8x}{64}+\frac{\sin 4x}{8}+C$