Question

How do you find the integral of $\int {\sin }^{3}xdx$ ?

Answer 1

Compute the required integral.

$$\begin{gathered} \int {\sin }^{3}xdx=\int \sin x(1-{\cos }^{2}x)dx\left[\because {\sin }^{2}x=1-{\cos }^{2}x\right] \\ \Rightarrow \int {\sin }^{3}xdx=\int \sin xdx-\int \sin x.{\cos }^{2}xdx \end{gathered}$$

Now, solve the first integral,

$\int \sin xdx=\cos x+\hspace{0.17em}C$

Now, solve the second integral,

let us assume $\cos x=u$

therefore, $-\sin xdx=du$

substitute the value,

$$\begin{gathered} -\int \sin x.{\cos }^{2}xdx=\int u^{2}du \\ \Rightarrow -\int \sin x.{\cos }^{2}xdx=\frac{u^3}{3}+C \\ \Rightarrow -\int \sin x.{\cos }^{2}xdx=\frac{1}{3}{\cos }^{3}x+C \end{gathered}$$

Combining all the above,$\int {\sin }^{3}xdx=\cos x+\frac{1}{3}{\cos }^{3}x+C$

Hence the required value of $\int {\sin }^{3}xdx$ is $\cos x+\frac{1}{3}{\cos }^{3}x+C$