Answer
$$\int^{\pi/2}_0\sin^22\theta\cos^32\theta d\theta=0$$
Work Step by Step
$$A=\int^{\pi/2}_0\sin^22\theta\cos^32\theta d\theta$$ $$A=\frac{1}{2}\int^{\pi/2}_0\sin^22\theta\cos^32\theta d(2\theta)$$
We set $a=2\theta$.
For $\theta=\pi/2$, $a=\pi$ and for $\theta=0$, $a=0$
$$A=\frac{1}{2}\int^\pi_0\sin^2a\cos^3ada$$
This is Case 2, so we would rewrite $$\cos^3a=\cos^2a\cos a=(1-\sin^2a)\cos a$$
Therefore,
$$A=\frac{1}{2}\int^\pi_0\sin^2a(1-\sin^2a)\cos ada$$ $$A=\frac{1}{2}\int^\pi_0(\sin^2a-\sin^4a)d(\sin a)$$
Take $u=\sin a$
For $a=\pi$, $u=\sin\pi=0$ and for $a=0$, $u=0$
Therefore,
$$A=\frac{1}{2}\int^0_0(u^2-u^4)du=0$$
