Answer
$$\int^{\pi/2}_05(\sin x)^{3/2}\cos xdx=2$$
Work Step by Step
$$A=\int^{\pi/2}_05(\sin x)^{3/2}\cos xdx$$
We set $u=\sin x$, which means $$du=\cos xdx$$
For $x=\pi/2$, $u=1$ and for $x=0$, $u=0$
Therefore, $$A=5\int^1_0u^{3/2}du$$ $$A=\Big(\frac{5u^{5/2}}{\frac{5}{2}}\Big)\Big]^1_0=2u^{5/2}\Big]^1_0$$ $$A=2(1^{5/2}-0)$$ $$A=2\times1=2$$
