How can I evaluate $$\int\limits_C(1+x^2y)\ ds$$ where $C$ is the first quarter of the ellipse $\frac{x^2}9+\frac{y^2}4=1$
I tried parameterizing the curve but I couldn't get rid of the square root $$\sqrt{9\sin^2t +4\cos^2t}$$ The last integral I have reached is $$\int_0^{\pi/2}(1+18\cos^2t\sin t)\sqrt{9\sin^2t +4\cos^2t} \ dt$$
On
There is a contradiction in the wording of the question.
Either $$\int\limits_C(1+x^2y)\ ds = \int_0^{\pi/2}(1+18\cos^2(t)\sin(t))\sqrt{9\sin^2(t) +4\cos^2(t)} \ dt$$ or $$\int\limits_C(1+x^2y^2)\ ds = \int_0^{\pi/2}(1+36\cos^2(t)\sin(t)^2)\sqrt{9\sin^2(t) +4\cos^2(t)} \ dt$$ Please clarify which one you intend to solve.
Or equivalently, by elimination of $x$ with $x^2=9-\frac94 y^2$ : $$\int\limits_C(1+x^2y^2)\ ds = \frac18 \int_0^2(4+36y^2-9y^4)\sqrt{\frac{5y^2+16}{4-y^2}}dy$$ A closed form involves the Complete Elliptic Integral : http://mathworld.wolfram.com/EllipticIntegral.html
Not an answer, but merely a hint
You might want to try substituting $u = \sin t$, so that $u^2 = \sin^2 t$, and $1-u^2 = \cos^2 t$. Of course, you have $du= \cos t dt$, but you can write this as $$du = \sqrt{1 - u^2} dt$$ i.e. $$ dt = \frac{1}{\sqrt{1-u^2}} du $$ and then you'll end up with an expression involving polynomials and square roots of a polynomial. But the polynomial will be a function of $u^2$ rather than $u$, which gives some small hope for further progress. I suspect an arctan will enter in somewhere. But who knows?