Actually, I think the first problem is in the limits for t. As you go from (0,2) to (-2,0), that is, with $\displaystyle c(t)=(2\cos{t},2\sin{t})$, then $\displaystyle \frac{\pi}{2}\le{t}\le\pi$.

Next $\displaystyle x(t)=2\cos{t}$, and $\displaystyle y(t)=2\sin{t}$, so $\displaystyle f(c(t))=xy=4\sin{t}\cos{t}$.

And since this is

the line integral of a scalar function, ds is the arc length parameter:

$\displaystyle ds^2=dx^2+dy^2$, so in terms of t, $\displaystyle ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{ dy}{dt}\right)^2}\,dt=\sqrt{(-2\sin{t})^2+(2\cos{t})^2}\,dt$

$\displaystyle ds=\sqrt{4\sin^2{t}+4\cos^2{t}}\,dt=2\,dt$

and you have

$\displaystyle \int_{\pi/2}^{\pi}(4\sin{t}\cos{t})2dt=8\int_{\pi/2}^{\pi}\sin{t}\cos{t}\,dt$.

--Kevin C.