Solving differential equation describing motion in a pendulum

I've been looking at Simple Harmonic Motion in particularly the period of a pendulum. This may seem like physics but my question is tailored towards mathematics. The differential equation is:

$${{d^2\theta}\over dt^2}+\sin\theta=0$$

Using the small angle approximation it is found that $T={2\pi}\sqrt{L\over g}$.

Is it possible to solve the differential equation without using the small angle approximation? If so, what is the actual period of a pendulum?

• I don't think there's a nice expression for the trajectory but there's a decently nice formula for the period: en.wikipedia.org/wiki/… – James Hanson Jan 31 '16 at 6:38
• I tried to find the expression for the trajectory, I believe it uses elliptical integrals. – Nikunj Jan 31 '16 at 6:45
• Yeah essentially it will all boil down to elliptic integrals. – Triatticus Jan 31 '16 at 7:18

Let $\ell=$ length measured to the C.G. of the bob, $I=$ moment of inertia of the bob about the end of the string and $E=$ total energy.

\begin{align*} \ddot{\theta}+\omega^{2} \sin \theta &= 0 \\ \omega &= \sqrt{\frac{mg\ell}{I}} \\ k &= \sqrt{\frac{E}{2mg \ell}} \end{align*}

$$\begin{array}{|c|c|c|c|} \hline & k < 1 & k = 1 & k > 1 \\ \hline & & & \\ \displaystyle \sin \frac{\theta}{2} & k\operatorname{sn} (\omega t,k) & \tanh \omega t & \displaystyle \operatorname{sn} \left( k\omega t,\frac{1}{k} \right) \\ & & &\\ \theta & 2\sin^{-1} (k\operatorname{sn} (\omega t,k)) & 4\tan^{-1} e^{\omega t}-\pi & \displaystyle 2\operatorname{am} \left( k\omega t,\frac{1}{k} \right) \\ & & &\\ T & \displaystyle \frac{4K(k)}{\omega} & \infty & \displaystyle \frac{2K(\frac{1}{k})}{k\omega} \\ & & &\\ \hline \end{array}$$

For small bob, $I\approx m\ell^{2}$.

For $k<1$, amplitude $\alpha=2\sin^{-1} k$ and $k>1$ it's moving in complete circle.

For $k<<1$, $T=2\pi \sqrt{\frac{I}{m g\ell}} \left( 1+\frac{1}{4} \sin^{2} \frac{\alpha}{2}+\ldots \right)$

A plot of $T$ vs. $k$ with $\omega=1$ is shown below