Answer
$$-\frac{1}{60} \cos^{10}6\theta +\frac{1}{72} \cos^{12}6\theta+C$$
Work Step by Step
\begin{aligned} \int \cos ^{9} 6 \theta \sin ^{3} 6 \theta d \theta &=\int \cos ^{9} 6 \theta \sin ^{2} 6 \theta \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta\left(1-\cos ^{2} 6 \theta\right) \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta \sin 6 \theta-\cos ^{11} 6 \theta \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta \sin 6 \theta d \theta-\int \cos ^{11} 6 \theta \sin 6 \theta d \theta \\
&= -\frac{1}{60} \cos^{10}6\theta +\frac{1}{72} \cos^{12}6\theta+C
\end{aligned}