Answer
$$\frac{1}{6}\sin ^{6} \theta -\frac{1}{8}\sin ^{8} \theta +C$$
Work Step by Step
\begin{aligned} \int \sin ^{5} \theta \cos ^{3} \theta d \theta &=\int \sin ^{5} \theta \cos ^{2} \theta \cos \theta d \theta \\ &=\int \sin ^{5} \theta\left(1-\sin ^{2} \theta\right) \cos \theta d \theta \\ &=\int \sin ^{5} \theta \cos \theta-\sin ^{7} \theta \cos \theta d \theta\\
&=\frac{1}{6}\sin ^{6} \theta -\frac{1}{8}\sin ^{8} \theta +C
\end{aligned}
