Answer
\begin{aligned}
\int \sin ^{3} x d x
=-\cos x+\frac{1}{3} \cos ^{3} x+C
\end{aligned}
Work Step by Step
Given $$\begin{equation}
\int \sin ^{3} x d x
\end{equation}$$
So, we have
\begin{aligned}
I&=
\int \sin ^{3} x d x\\
&=\int \sin ^{2} x \sin x d x\\
&=\int\left(1-\cos ^{2} x\right) \sin x d x\\
&=\int \sin x d x-\int \cos ^{2} x \sin x d x\\
&=-\cos x+\frac{1}{3} \cos ^{3} x+C
\end{aligned}
