## Thomas' Calculus 13th Edition

$$\int_{0}^{\pi / 2} \sin ^{2} 2 \theta \cos ^{3} 2 \theta \ d\theta=0$$
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}