Question

What is the integral of $si{n}^4\left(x\right)$?

Answer 1

Find the integral of $si{n}^4\left(x\right)$

Reframe and further calculate the integral:

$\begin{array}{rcl}\int {\sin }^{4}xdx& =& \int {\left({\sin }^{2}x\right)}^{2}dx\\ & \Rightarrow & \int {\sin }^{4}xdx=\int {\left(\frac{1-\cos 2x}{2}\right)}^{2}dx\left\{\because si{n}^{2}x=\frac{1-cos2x}{2}\right\}\\ & \Rightarrow & \int {\sin }^{4}xdx=\frac{1}{4}\int 1+{\cos }^{2}2x-2\cos 2xdx\\ & \Rightarrow & \int {\sin }^{4}xdx=\frac{1}{4}\int 1+\left(\frac{1+\cos 4x}{2}\right)-2\cos 2xdx\left\{\because co{s}^{2}2x=\frac{1+cos4x}{2}\right\}\\ & \Rightarrow & \int {\sin }^{4}xdx=\frac{1}{4}\left[\frac{3x}{2}+\frac{\sin 4x}{8}-\frac{2\sin 2x}{2}\right]+C\left\{\because \int \cos \theta =\sin \theta .\mathrm{Also},C\mathrm{is}a\mathrm{constant}\right\}\\ & \Rightarrow & \int {\sin }^{4}xdx=\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}+C\end{array}$

Hence, $\begin{array}{rcl}\int {\sin }^{4}xdx& =& \frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}+C\end{array}$