Answer
\begin{aligned}
\int \sin ^{5} x d x =-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C
\end{aligned}
Work Step by Step
Given $$ \int \sin ^{5} x d x $$
So, we have
\begin{aligned}
I&= \int \sin ^{5} x d x\\
&=\int\left(\sin ^{2} x\right)^{2} \sin x d x\\
&=\int\left(1-\cos ^{2} x\right)^{2} \sin x d x\\
&=\int\left(1-2 \cos ^{2} x+\cos ^{4} x\right) \sin x d x\\
&=\int \sin x d x-\int 2 \cos ^{2} x \sin x d x+\int \cos ^{4} x \sin x d x\\
&=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C
\end{aligned}
