Answer
$$ \int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos ^{3} 2 \theta \ d\theta=0 $$
Work Step by Step
Given $$ \int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos ^{3} 2 \theta \ d\theta $$
So, we have
\begin{aligned}
I&= \int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos ^{3} 2 \theta d \theta\\
&= \int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos ^{2} 2 \theta \cos 2 \theta d \theta\\
&=\int_{0}^{\pi / 2} \sin ^{2} 2 \theta\left(1-\sin ^{2} 2 \theta\right) \cos 2 \theta \ d \theta\\
&=\int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos 2 \theta \ d \theta-\int_{0}^{\pi / 2} \sin ^{4} 2 \theta \cos 2 \theta \ d \theta\\
&=\frac{1}{2}\int_{0}^{\pi / 2} \sin ^{2} 2 \theta\ \ ( 2\cos 2) \theta \ d \theta-\frac{1}{2}\int_{0}^{\pi / 2} \sin ^{4} 2 \theta (2\cos 2 \theta) \ d \theta\\
&=\left[\frac{1}{2} \cdot \frac{\sin ^{3} 2 \theta}{3}-\frac{1}{2} \cdot \frac{\sin ^{5} 2 \theta}{5}\right]_{0}^{\pi / 2}\\
&=\left[\frac{1}{2} \cdot \frac{\sin ^{3} \pi}{3}-\frac{1}{2} \cdot \frac{\sin ^{5} \pi}{5}\right]-\left[\frac{1}{2} \cdot \frac{\sin ^{3} 0}{3}-\frac{1}{2} \cdot \frac{\sin ^{5} 0}{5}\right]\\
&=0
\end{aligned}
