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I've been reading through the math on nonlinear (large amplitude) pendulum in wikipedia, and it beats me on how can

$\displaystyle\int_{0}^{\theta_0} \dfrac{d\theta}{\sqrt{\cos(\theta) - \cos(\theta_0)}}$

re-written into this elliptic integral term;

$F \left [ \dfrac{\theta_0}{2},{\csc\left(\dfrac{\theta_0}{2}\right)} \right ]\csc \left (\dfrac{\theta_0}{2} \right )$

Any helpful guidances will be appreciated.

update:

Here is my work so far,

from here;

$$\displaystyle\int_{0}^{\theta_0} \dfrac{d\theta}{\sqrt{\cos(\theta) - cos(\theta_0)}}$$

I used the half-angle formula $\cos\left( \theta \right) = 1-2 \sin^2\left( \frac{\theta}{2} \right)$,

$$\require{cancel}\displaystyle\int_{0}^{\theta_0} \dfrac{d\theta}{\sqrt{\cancel1- 2 \sin^2\left( \frac{\theta}{2} \right) \cancel{-1} + 2 \sin^2\left( \frac{\theta_0}{2} \right)}}$$

Then, multiply both numerator and denominator with $\csc(\frac{\theta_0}{2})$,

$$\displaystyle\int_{0}^{\theta_0} \dfrac{\csc\left(\frac{\theta_0}{2}\right)d\theta}{\sqrt{2 - 2\csc^2\left( \frac{\theta_0}{2} \right)\sin^2\left(\frac{\theta}{2}\right)}}$$

Almost there, but how the coefficient of 2 in front of $2\csc^2\left( \frac{\theta_0}{2} \right)\sin^2\left(\frac{\theta}{2}\right)$ is eliminated in the process to got the $F$ function as said above?

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