<s>[INST] Integral Bounds of Part of an Ellipse 
I'm looking through my notes, and I've completed this questions integral, getting $ -144 \int cos^4(\theta)$ and due to the fact that $x=4cos(\theta)$ I calculated that the lower bound is $cos^-1(\frac{2\sqrt{3}}{4})=\frac{\pi}{6}$. The lecture notes confirm that this is correct however when I try the upper bound of the integral I get two different answers regardless of how I use $x=4cos(\theta)$ or $y=6sin(\theta)$ (the two ways in which he curve is parametererised).
The upper bound that the lecture notes specify is $\frac{4\pi}{3}$
 [/INST] You need to solve both equations at once.
$$
    -2 = 4 \cos\theta\qquad\qquad
    -3  \sqrt 3 = 6 \sin\theta
$$
If you try to use inverse functions on the first, you get an angle in the second quadrant: $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.  If you try to use inverse functions on the second, you get an angle in the fourth quadrant: $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$.  Neither of these is the angle you are looking for: the point is in the third quadrant.
However, using the symmetry properties of sine and cosine, we see that:
\begin{align*}
   \cos\left(-\frac{2\pi}{3}\right) &= \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\\
   \sin\left(-\frac{2\pi}{3}\right) &= -\sin\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}
\end{align*}
So this angle satisfies both equations.  But we need to start from the lower limit $\theta=\frac{\pi}{6}$ and go counterclockwise, increasing $\theta$.  So we use $-\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}$.
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