Question

Find the integral of the function
${\sin }^{4}x$

Answer 1

Consider the given integral.

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

$I=\int {\left({\sin }^{2}x\right)}^{2}dx$

We know that

$\cos 2x=1-2{\sin }^{2}x$

${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$

Therefore,

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

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

We know that

$\cos 2\left(2x\right)=2{\cos }^{2}2x-1$

${\cos }^{2}2x=\frac{1}{2}(1+\cos 2\left(2x\right))$

Therefore,

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

$I=\frac{1}{4}[\int \frac{3}{2}dx+\frac{1}{2}\int \cos 4xdx-2\int \cos 2xdx]$

$I=\frac{1}{4}[\frac{3x}{2}+\frac{1}{2}\left(\frac{\sin 4x}{4}\right)-2\left(\frac{\sin 2x}{2}\right)]+C$

$I=\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}+C$

Hence, this is the answer.