After differentiating and doing some algebra, one arrives at the following DE:
d2theta/dt2 + (mgL/I) sin(theta) = 0
The period can be found by integrating the dt/dtheta expression above, performing a change of integration variable, and using a trig half-angle formula. The result is a complete elliptic integral, K(k):
T = 4 sqrt(I/mgL) K(sin(theta0/2))
Plotting the period, T, as a function of starting angle (assuming dtheta/dt=0 at theta0):
Near theta0=0, one finds the solution (T0~2s) for the small-angle approximation: sin(theta) ~ theta.
The period of the swinging-body simulation is found to match the calculation of the period from the max angle:
Keep in mind that this acts as a check for the former, not the latter.