Question

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

Answer 1

${\sin }^{4}x={\sin }^{2}x{\sin }^{2}x$
$=\left(\frac{1-\cos 2x}{2}\right)\left(\frac{1-\cos 2x}{2}\right)$
$=\frac{1}{4}(1-\cos 2x{)}^2$
$=\frac{1}{4}[1+{\cos }^{2}2x-2\cos 2x]$
$=\frac{1}{4}[1+\left(\frac{1+\cos 4x}{2}\right)-2\cos 2x]$
$=\frac{1}{4}[1+\frac{1}{2}+\frac{1}{2}\cos 4x-2\cos 2x]$
$=\frac{1}{4}[\frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x]$
$\therefore \int {\sin }^{4}xdx=\frac{1}{4}\int [\frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x]dx$
$=\frac{1}{4}[\frac{3}{2}x+\frac{1}{2}\left(\frac{\sin 4x}{4}\right)-\frac{2\sin 2x}{2}]+C$
$=\frac{1}{8}[3x+\frac{\sin 4x}{4}-2\sin 2x]+C$
$=\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C$