Answer
$$\int \sin ^{5} x d x = \frac{-1}{5}\sin ^{4} x \cos x-\frac{4}{15} \sin ^{2} x \cos x-\frac{8}{15} \cos x +C$$
Work Step by Step
$$
\int \sin ^{5} x d x
$$
Here $n=5$, using the formula
$$
\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x
$$
We get
\begin{align*}
\int \sin ^{5} x d x&=-\frac{\sin ^{4} x \cos x}{5}+\frac{4}{5} \int \sin ^{3} x d x\\
&=-\frac{\sin ^{4} x \cos x}{5}+\frac{4}{5}\left(-\frac{\sin ^{2} x \cos x}{3}+\frac{2}{3} \int \sin x d x\right)\\
&=-\frac{\sin ^{4} x \cos x}{5}+\frac{4}{5}\left(-\frac{\sin ^{2} x \cos x}{3}-\frac{2}{3} \cos x \right)+C\\
&= \frac{-1}{5}\sin ^{4} x \cos x-\frac{4}{15} \sin ^{2} x \cos x-\frac{8}{15} \cos x +C
\end{align*}
