<s>[INST] Line integral over ellipse in first quadrant 
Evaluate $ \int_{C} xy\,ds $ where C is the arc of the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ in the first quadrant.

Let $x = a\cos t$ and $ y= b\sin t$ and use a substitution of $ u = a^2 \sin^2t + b^2\cos^2t $ to simply the expression under the sqrt root when dealing with $ ds$.
I eventually get to $$\frac{ab}{2(a^2-b^2)} [\frac{2}{3} \sqrt{(a^2\sin^2t + b^2 \cos^2t)^3}]$$evaluated between $\pi/2$ and $0$. 
Should I take $\sqrt{a^6} = a^3$ here? (my thoughts being that $ a>0 $  in first quadrant)
 [/INST] The correct setup is
$$\int_{0}^{2 \pi} ab\cos(t) \sin(t) \sqrt{a^2 \cos^2t+b^2\sin^2t}dt$$
As you hypothesized. Let $u = a^2 \cos^2t+b^2\sin^2t$ and $du = 2b^2\cos(t)\sin(t) - 2a^2\sin(t) \cos(t) dt$ 
Factor out $\cos(t) \sin(t)$, this will cancel out the same factor upstairs and in the end you should end up with
$$\int \dfrac{ab \sqrt{u} du}{2b^2 - 2a^2}$$
Factor out the constants and integrate. Then do the definite integral (which is very similar).
I leave the rest to you
</s>