Question

Integrate: ${\sin }^{4}x$

Answer 1

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