Question

Integrate $\int {\sin }^3\left(\mathrm{x}\right)dx$.

Answer 1

Compute the given integration.

In the question, an integral $\int {\sin }^3\left(\mathrm{x}\right)dx$ is given.

We know that,

$$\begin{gathered} \sin \left(3\mathrm{x}\right)=3\sin \left(\mathrm{x}\right)-4{\sin }^3\left(\mathrm{x}\right) \\ \Rightarrow {\sin }^3\left(\mathrm{x}\right)=\frac{3\sin \left(\mathrm{x}\right)-\sin \left(3\mathrm{x}\right)}{4} \end{gathered}$$

Assume that, $I=\int {\sin }^3\left(\mathrm{x}\right)\mathrm{dx}$.

So,

$$\begin{gathered} I=\int \frac{3\sin \left(\mathrm{x}\right)-\sin \left(3\mathrm{x}\right)}{4}\mathrm{dx} \\ \Rightarrow I=\frac{3}{4}\int \sin \left(\mathrm{x}\right)-\frac{1}{4}\int \sin \left(3\mathrm{x}\right)\left[\because \int \sin \left(\mathrm{x}\right)=-\mathrm{cosx},\int \sin \left(\mathrm{cx}\right)=\frac{-\cos \left(\mathrm{cx}\right)}{\mathrm{c}}\right] \\ \Rightarrow I=\frac{-3}{4}\cos \left(\mathrm{x}\right)+\frac{\cos \left(3\mathrm{x}\right)}{12}+\mathrm{C} \end{gathered}$$

Hence $\int {\sin }^3\left(\mathrm{x}\right)dx=\frac{-3}{4}\cos \left(\mathrm{x}\right)+\frac{\cos \left(3\mathrm{x}\right)}{12}+\mathrm{C}$.