Answer
$$-\frac{4}{5} \cos ^{5} \theta+\frac{4}{3} \cos ^{3} \theta-\cos \theta+C $$
Work Step by Step
We integrate as follows:
\begin{align*}
\int \cos ^{2} 2 \theta \sin \theta d \theta & =\int\left(2 \cos ^{2} \theta-1\right)^{2} \sin \theta d \theta\\
&=\int\left(4 \cos ^{4} \theta-4 \cos ^{2} \theta+1\right) \sin \theta d \theta\\
&=\int 4 \cos ^{4} \theta \sin \theta d \theta-\int 4 \cos ^{2} \theta \sin \theta d \theta+\int \sin \theta d \theta\\
&=-\frac{4}{5} \cos ^{5} \theta+\frac{4}{3} \cos ^{3} \theta-\cos \theta+C
\end{align*}
