Question

Evaluate :
$\int si{n}^{4}x$ dx.

Answer 1

$=\int {\sin }^{4}xdx$

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

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

$=\frac{1}{4}\int dx+\frac{1}{4}\int {\cos }^{2}2xdx-\frac{2}{4}\int \cos 2xdx$

$=\frac{1}{4}x+\frac{1}{8}\int 2{\cos }^{2}2xdx-\frac{1}{2}\frac{\sin 2x}{2}+c$

$=\frac{x}{4}+\frac{1}{8}\int (1+\cos 4x)dx-\frac{1}{2}\frac{\sin 2x}{2}+c$

$=\frac{x}{4}+\frac{x}{8}+\frac{\cos 4x}{32}-\frac{1}{2}\frac{\sin 2x}{2}+c$

$=\frac{2x+x}{8}+\frac{\cos 4x}{32}-\frac{\sin 2x}{4}+c$

$=\frac{3x}{8}+\frac{\cos 4x}{32}-\frac{\sin 2x}{4}+c$