Answer
$$\int^{\pi}_03\cos^2 x\sin xdx=\int^{3\pi}_{2\pi}3\cos^2 x\sin xdx=2$$
Work Step by Step
Part a) $$A=\int^{\pi}_03\cos^2 x\sin xdx$$
Set $u=\cos x$, we then have $$du=-\sin xdx$$ $$\sin xdx=-du$$
For $x=0$, we have $u=1$ and for $x=\pi$, we have $u=-1$.
Therefore, $$A=-\int^{-1}_13u^2du$$ $$A=-\frac{3u^3}{3}\Big]^{-1}_1=-u^3\Big]^{-1}_1$$ $$A=-(-1-1)=2$$
Part b) $$A=\int^{3\pi}_{2\pi}3\cos^2 x\sin xdx$$
We carry out a similar substitution as in part a).
For $x=2\pi$, we have $u=0$ and for $x=3\pi$, we have $u=-1$.
Therefore, $$A=-\int^{-1}_{1}3u^2du=2$$
