Answer
\begin{aligned}
\int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta =-\cos ^{4} 2 \theta+C
\end{aligned}
Work Step by Step
Given $$ \int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta $$
So, we have
\begin{aligned}
I&=\int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta\\
&=8\int (-\frac{1}{2}) (\cos 2 \theta )^3(-2\sin 2 \theta) \ d \theta\\
&=-4\frac{\cos ^{4} 2 \theta}{4}+C\\
&=-\cos ^{4} 2 \theta+C
\end{aligned}
