Solving a curve integral around part of an elipse I'm having trouble calculating a curve integral in a vector field:
$\int_C  y (18x + 1)\ dx +  2y^2\ dy$ 
where $C$ is the curve along the ellipse
$9x^2 +  y^2 = 64$ 
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3} $ , $4$) to the point (-$\frac{4}{3}$ , $4\sqrt{3} $)
Thats almost one "lap" around the ellipse.. 
When making a parametrisation I come up with:
x = $\sqrt{ \frac{64}{9}\ }  \cos t $
y = $\sqrt{ 64 }  \sin t $
$- \frac{\Pi}{3} \le t < \arctan(3  \sqrt{3}) $
But the integral created from this parametrization give an answer involving arctan. Any ideas to get a rational answer?
I had an idea to split the curve into multiple curves and integrating them piecewise, but that gets really messy aswell.
This is my calculations when making a variable change:
First the variable change:
$  u= 3x $ 
$ du = 3dx $
$ \ u^2 + y^2 =64 $
$ \int y(18x + 1)dx + 2y^2dy = \int y(6u + 1)\frac{du}{3} + 2y^2dy $
$ u = \sqrt{64}\cos t $ 
$ y = \sqrt{64}\sin t $ 
$ du = -\sqrt{64}\sin tdt $ 
$ dy = \sqrt{64}\cos tdt $ 
The startpoint after making variable change:
$ ( -4\sqrt{3},4)$ 
$ arctan( \frac{4}{-4\sqrt{3}}) = -\Pi/6$ 
Startangle:  $ \frac{5\Pi}{6} $
The endpoint after making variable change: $ ( -4,4\sqrt{3})$
$ arctan( \frac{-4\sqrt{3}}{4}) = -\Pi/3$ 
End angle: $\frac{2\Pi}{3} $
$\int y(6u + 1)\frac{du}{3} + 2y^2dy = \int -\frac{1}{3}\sqrt{64}\sin t(6\sqrt{64}\cos t + 1)\sqrt{64}\sin tdt + 128\sin^2t\sqrt{64}\cos tdt =  $
$\int -\frac{64}{3}\sin^2tdt $
$ \frac{5\Pi}{6} \le  t \le  \frac{2\Pi}{3} $
The answer becomes: $\frac{16}{9}\Pi $
 A: Draw a figure! The parametrization of the ellipse is
$$t\mapsto \bigl(x(t),y(t)\bigr):=\left({8\over3}\cos t,8\sin t\right)\qquad(t\in{\mathbb R})\ .\tag{1}$$
Now we have to find the boundary values which are relevant for the integral $J$ in question. Both endpoints $p$ and $q$ of the arc lie in the second quadrant. At the starting point we have $8\sin t=4$, or $\sin t={1\over2}$, which amounts to $t=-\pi-{\pi\over6}$, and at the endpoint we have $8\sin t=4\sqrt{3}$, which amounts to $t={2\pi\over3}$.
Therefore we  obtain
$$J=\int_{-7\pi/6}^{2\pi/3} \left(y(t)(18 x(t)+1)\dot x(t) + 2y^2(t)\dot y(t)\right)\ dt\ ,$$
where you now have to plug in the parametrization $(1)$. The computation can be simplified somewhat by observing that
$$\int\nolimits_\gamma 2y^2\ dy={2\over3}y^3\biggr|_p^q\quad .$$
A: As @dfan suggested, substitute $u=3x$ leading to a circular curve $C'$ defined by $$u^2+y^2=r^2$$ of radius $r=8$. Your integral becomes
$$J:=\int_C dx\,y(18x+1)+\int_C dy\,2y^2=\frac{1}{3}\int_{C'}du\,y(6u+1)+2\int_{C'}dy\,y^2.$$ Taking the total differential of the circle equation, we have $udu=-ydy$, so the second integral cancels against the $6u$-term from the first one and we are left with
$$J=\frac{1}{3}\int_{C'}du\,y.$$
Next, introduce polar coordinates
$$u=r\cos\phi,\\y=r\sin\phi,$$parametrizing the curve with $\phi$. Your integral limits transform to
$$(x,y)_0=(-4/\sqrt{3},4)\to u_0=-4\sqrt{3}\quad \to\phi_0=\arccos\frac{-\sqrt{3}}{2}=\frac{5\pi}{6},$$ 
$$(x,y)_1=(-4/3,4\sqrt{3})\to u_1=-4\quad \to\phi_1=\arccos\frac{-1}{2}+2\pi=\frac{8\pi}{3}.$$ 
Here it is essential to add $2\pi$ to the upper limit $\phi_1$, such that $\phi_0<\phi_1$ and the integral passes the curve in the correct rotation sense. Finally, with the line element
$$du=-d\phi\,r\sin\phi,$$ the integral becomes
$$J=-\frac{r^2}{3}\int_{\phi_0}^{\phi_1} d\phi\,\sin^2\phi=-\frac{r^2}{6}\left( \phi-\frac{1}{2}\sin2\phi \right)_{\phi_0}^{\phi_1}$$ and using the given values for $r,\phi_0,\phi_1$ we obtain
$$J=-\frac{176\pi}{9}.$$
