Question

Solve:- $\int {\sin }^{4}xdx$

Answer 1

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

$=\frac{1}{4}\int {\left(2{\sin }^{2}x\right)}^{2}dx$

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

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

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

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

$=\frac{1}{4}\int dx+\frac{1}{8}\int \cos 4xdx+\frac{1}{8}\int dx-\frac{2}{4}\int \cos 2xdx$

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

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

$=\frac{3}{8}x+\frac{1}{32}\sin 4x-\frac{1}{4}\sin 2x+c$