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? 
 A: 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

