{1}{k_1}\sqrt{\frac{2I}{(E+mgl)}} \int_{\phi(0)}^{\phi(t)}
\frac{d\phi}{\sqrt{1 - \sin^2(\phi)}} \frac{\cos(\phi)}{\sqrt{1 -
\frac{\sin^2\phi}{k_1^2}}} \\ = \sqrt{\frac{I}{mgl}} \left[
F\left(\phi(t),k_2\right) - F\left(\phi(0),k_2\right) \right],\quad
k_2 = \frac{1}{k_1}.\end{gather*} The inverse of $F$ is given by the
Jacobi elliptic functions (sn,cn,...), yielding:
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\begin{gather*}\sin(\phi(t)) = \text{sn}\left(t
  \sqrt{\frac{mgl}{I