(surface) Area of an ellipse by integrating Given is an ellipse with $x=a\cos(t),~~y=b\sin(t)$
I do this by using $S=|\int_c^d x(t)y'(t) dt|$, so calculating the area regarding the vertical axis. 
Since $t$ runs from $0$ to $2\pi$ I figured I only had to calculate it from $c=\pi/2$ to $d=3\pi/2$ and then this times $2$. But when I integrate over those I get zero...
My steps:
\begin{align*}
0.5S & = |\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d ab\cos^2(t) dt|\\
S & = 2ab|\int_c^d 2\cos(t)*(-\sin(t)) dt|\\
S & = 4ab|\int_c^d -\cos(t)\sin(t) dt|
\end{align*}
On those bounds $\cos(t)$ is zero, so how will this work?!
I notice that to integrate $\cos^2(x)$, most use $\cos^2x = 1/2\cos(2x) + 1/2 $ and thus find $1/4 sin(2x) + 1/2 x$. 
This works for those boundaries (except for a factor 2 somewhere ???), but how do I know NOT to use $\cos(x)\sin(x)$ ?!

I think the integral must be a line integral, covering the ellipse! The area of the ellipse is $\int\limits_{0}^{2 \pi} x(t)y'(t)dt|=|\int\limits_{0}^{2\pi}ab \cos^{2}t dt|=|ab| \int\limits_{0}^{2\pi} \frac{1+cos(2t)}{2}dt=|ab| \pi$
