How Do You Derive the Period of a Pendulum with Arbitrary Amplitude?

[StrangelyQuarky]

Homework Statement

A pendulum obeys the equation \ddot{\theta} = -\sin(\theta) and has amplitude \theta_0. I have to show that the period is
T = 4 \int_{0}^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-\alpha \sin^2(\phi)}} where \alpha = \sin^2(\frac{\theta_0}{2})

2. The attempt at a solution

I derived an expression for time:

\dot{\theta}\frac{d\dot{\theta}}{d\theta} = -\sin(\theta)

I said that the pendulum starts out at the height of its amplitude \theta = \theta_0 where it also has zero velocity

\int_{0}^{\dot{\theta}} \dot{\theta}d\dot{\theta} = \int_{\theta_0}^{\theta} -\sin(\theta)d\theta

\dot{\theta} = \frac{d\theta}{dt}= \pm \sqrt{2(\cos(\theta)-\cos(\theta_0))}

So for the period we can integrate from \theta_0 \text{ to } 0, which is a quarter of the period, then multiply by 4 to get the whole period.

T = 4 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{2(\cos(\theta)-\cos(\theta_0))}}

By the half-angle identity,

T = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\sin^2(\theta_0 /2)-\sin^2(\theta /2)}} = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\alpha-\sin^2(\theta /2)}}

And this is where I'm stuck. It looks similar to the desired answer, but I can't think of any identity or substitution that would give the right integrand and limits.

 

[haruspex]

Take a look at https://en.m.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period