Differential equation for the simple pendulum

[Celso]

Homework Statement

A simple pendulum whose length is $l = 9.8m$ satisfies the equation $\dddot\theta + sin(\theta) = 0$
If $\Theta_{0}$ is the amplitude of oscillation, show that its period $T$ is given by
$T = 4 \int_{0}^{\frac{\pi}{2}} \frac{d\phi}{(1-\alpha sin(\phi)^2)^{1/2}}$ where $\alpha = sin(\frac{1}{2}\Theta)^2$

Relevant Equations

$T = \frac{2\pi}{\omega}$

How do I start this? I plugged the differential equation at wolfram alpha and it semmed too complicated for such an exercise. I've also looked at a sample of an answer on cheeg where the initial approach is to rewrite the equation as $\frac{d}{dt} (\frac{\dot\theta^2}{2}-cos(\theta)) = 0$
How is this right if the $\theta$ time dependence is not given?

 

[kuruman]

It would help if you had the correct equation, $\ddot \theta+\sin(\theta)=0$. Note that what you found in cheeg is incorrect, $\frac{d}{dt} (\frac{\dot\theta^2}{2}-cos(\theta)) =\ddot \theta+\dot \theta \sin(\theta)$ which is not the same as above. Whoever posted this on cheeg forgot the chain rule of differentiation.

I would recommend that you start with the energy conservation equation, separate variables ($\theta$ on one side and $t$ on the other) and then integrate.