Answer
$$\frac{1}{9} \sin ^{9} \theta-\frac{1}{11} \sin ^{11} \theta+C$$
Work Step by Step
\begin{aligned} \int \cos ^{3} \theta \sin ^{8} \theta d \theta &=\int \cos \theta \cos ^{2} \theta \sin ^{8} \theta d \theta \\ &=\int \cos \theta\left(1-\sin ^{2} \theta\right) \sin ^{8} \theta d \theta \\ &=\int\left(\cos \theta-\cos \theta \sin ^{2} \theta\right) \sin ^{8} \theta d \theta \\ &=\int\left(\cos \theta \sin ^{8} \theta-\cos \theta \sin ^{10} \theta\right) d \theta \\ &=\int \cos \theta \sin ^{8} \theta d \theta-\int \cos \theta \sin ^{10} \theta d \theta\\
&=\frac{1}{9} \sin ^{9} \theta-\frac{1}{11} \sin ^{11} \theta+C
\end{aligned}
