Answer
\begin{aligned}
\int \cos ^{3} 2 x \sin ^{5} 2 x d x =\frac{1}{12} \sin ^{6} 2 x-\frac{1}{16} \sin ^{8} 2 x+C
\end{aligned}
Work Step by Step
Given $$ \int \cos ^{3} 2 x \sin ^{5} 2 x d x $$
So, we have
\begin{aligned}
I&= \int \cos ^{3} 2 x \sin ^{5} 2 x d x\\
&=\frac{1}{2} \int \cos ^{3} 2 x \sin ^{5} 2 x \cdot 2 d x\\
&=\frac{1}{2} \int \cos 2 x \cos ^{2} 2 x \sin ^{5} 2 x \cdot 2 d x\\
&=\frac{1}{2} \int\left(1-\sin ^{2} 2 x\right) \sin ^{5} 2 x \cos 2 x \cdot 2 d x\\
&=\frac{1}{2} \int \sin ^{5} 2 x \cos 2 x \cdot 2 d x-\frac{1}{2} \int \sin ^{7} 2 x \cos 2 x \cdot 2 d x\\
&=\frac{1}{12} \sin ^{6} 2 x-\frac{1}{16} \sin ^{8} 2 x+C
\end{aligned}
