Question

Integral ${\sin }^{4}x$.

Answer 1

Find the integral

Given, ${\sin }^{4}x$

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

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

We know that,

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

Therefore, ${\left({\sin }^{2}x\right)}^2={\left(\frac{1-\cos 2x}{2}\right)}^2$

Now,

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

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

We know that,

$$\begin{gathered} \Rightarrow \cos 2x=2{\cos }^{2}x-1 \\ \Rightarrow 2\cos {}^{2}x=\cos 2x+1 \\ \mathrm{So},2{\cos }^{2}2x=\cos 4x+1 \\ \Rightarrow {\cos }^{2}2x=\frac{\cos 4x+1}{2} \end{gathered}$$

So,

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

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

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

Hence , integral of ${\sin }^{4}x$ is $\frac{3x}{8}-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+c$.