Question

What is the integration of ${\sin }^{4}x$?

Answer 1

Find the integration of ${\sin }^{4}x$.

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

Take square on both sides :

$\begin{array}{rcl}{\sin }^{4}x& =& {\left(\frac{1-\cos 2x}{2}\right)}^2\end{array}$

To find the integration of ${\sin }^{4}x$, put integration on both sides:

$$\begin{gathered} \begin{array}{rcl}\int {\sin }^{4}xdx& =& \int {\left(\frac{1-\cos 2x}{2}\right)}^{2}dx\\ & \Rightarrow & \int {\sin }^{4}xdx=\frac{1}{4}\int 1+{\cos }^{2}2x-2\cos 2xdx\end{array} \\ \Rightarrow \int {\sin }^{4}xdx=\frac{1}{4}\int \begin{array}{rcl}& & \left(1+\frac{1+\cos 4x}{2}-2\cos 2x\right)\end{array}dx\left[\begin{array}{c}\mathbf{}\mathbf{}\mathbf{\because }\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{2}\mathbf{x}}{\mathbf{2}}\\ \mathbf{\therefore }\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{2}\mathbf{x}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{4}\mathbf{x}}{\mathbf{2}}\end{array}\right] \\ \Rightarrow \int {\sin }^{4}xdx=\frac{1}{4}\left(x+\frac{x}{2}+\frac{\sin 4x}{8}-\frac{2\sin 2x}{2}\right)+C \end{gathered}$$

Hence, the required integral is $\int {\sin }^{4}xdx=\frac{1}{4}\left(x+\frac{x}{2}+\frac{\sin 4x}{8}-\frac{2\sin 2x}{2}\right)+C$.