classical mechanics - Correction to Period of a Pendulum

In one derivation of the corrected period of a pendulum, we started off like so:

The mass has a height $y$ given by $l(1-\cos \theta )$. $E = K + E \rightarrow \frac{1}{2}ml^2 \dot{\theta}^2 + mgl(1-\cos \theta)$

The next step introduces $\theta _0$, and I've got no idea where this came from.

$$\frac{1}{2}ml^2 \dot{\theta}^2 + mgl(1-\cos \theta)= mgl(1-\cos \theta _0)$$

Now we just solve for $\dot{\theta}$ and solve the DE.W I'm interested in the theta side of the equation.

$$\int \frac{d\theta}{\sqrt{\cos \theta - \cos \theta _0 }}$$

We go through a bunch of subs and changes of vairble to arrive at

$$\int ^{2\pi} _0 \frac{du}{\sqrt{1-K^2 \sin ^2 u }}$$

So my two questions are

• Why are we involving two $\theta$ values? The text didn't make it clear why we needed an extra $\theta _0$. It appears that $mgl(1-\cos \theta _0)$ is the total energy of the system. Our total energy cannot surpass the initial gravitational potential energy, this is clear. My only thought as to what $\theta$ means is the instantaneous position of the angle.


• My text mentioned that this is an elliptical integral. Mathematically speaking, what is an elliptical integral? Can this integral be solved exactly, or does it always require approximations from the expansion?