Question

Solve:
$\int {\sin }^{4}x$

Answer 1

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

$I=\frac{1}{4}\int (1-2\cos 2x)dx\frac{1}{8}\int (1+\cos 4x)dx=\frac{x}{4}-\frac{2\sin 2x}{8}+\frac{x}{8}+\frac{\sin 4x}{32}+C=\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$