Answer
$$-\frac{1}{5} \sin ^{4} x \cos x-\frac{4}{15} \sin ^{2} x \cos x-\frac{8}{15} \cos x+C$$
Work Step by Step
Given $$\int \sin ^{5} x d x $$
By using
$$\int \sin ^{n} x d x=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x$$
With ($n=5$)
\begin{align*}
\int \sin ^{5} x d x&=-\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5} \int \sin ^{3} x d x\\
&= -\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5}\left(-\frac{1}{3} \sin ^{2} x \cos x+\frac{2}{3} \int \sin x d x\right)\\
&= -\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5}\left(-\frac{1}{3} \sin ^{2} x \cos x-\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*}
