Studying the nonlinear pendulum, I took the following system \begin{equation} \begin{cases} \theta(0)=0\\ \left(\frac{d\theta}{dt}\right)_{t=0}=0\\ \left(\frac{d\theta}{dt}\right)^2=4\omega_0^2\left[\sin^2{\frac{\theta_0}{2}}-\sin^2{\frac{\theta}{2}}\right] \end{cases} \end{equation}
and I got the solution $$\theta(t)=2\arcsin\left\{\sin{\frac{\theta_0}{2}}\textrm{sn}\left[K\left(\sin^2{\frac{\theta_0}{2}}\right)-\omega_0t;\sin^2{\frac{\theta_0}{2}}\right]\right\}.$$ Reference: http://www.scielo.br/pdf/rbef/v29n4/a24v29n4.pdf
So, I tried to get the ODE doing the reverse path but I fail: \begin{align} \dot{\theta}(t)&=-2\omega_0\sin{\frac{\theta_0}{2}}\textrm{cn}\left[K\left(\sin^2{\frac{\theta_0}{2}}\right)-\omega_0t;\sin^2{\frac{\theta_0}{2}}\right]\\ \dot{\theta}^2(t)&=4\omega_0^2\left(\sin^2{\frac{\theta_0}{2}}-\sin^2{\frac{\theta_0}{2}}\textrm{sn}^2\left[K\left(\sin^2{\frac{\theta_0}{2}}\right)-\omega_0t;\sin^2{\frac{\theta_0}{2}}\right]\right) \end{align} and I get stuck in here. Is it true that $$\sin^2{\frac{\theta_0}{2}}\textrm{sn}^2\left[K\left(\sin^2{\frac{\theta_0}{2}}\right)-\omega_0t;\sin^2{\frac{\theta_0}{2}}\right]=\sin^2{\frac{\theta}{2}}?$$ If this is the case, why?
Yes, it's true that $$\sin^2{\frac{\theta_0}{2}}\textrm{sn}^2\left[K\left(\sin^2{\frac{\theta_0}{2}}\right)-\omega_0t;\sin^2{\frac{\theta_0}{2}}\right]=\sin^2{\frac{\theta}{2}}.$$
Because, following the notation established on the referred paper, $$F(\arcsin z; k)= K(k)-\tau,$$ where $F(\varphi,m)$ and $K(m)$ are the complete and the incomplete elliptic integral of the first kind, respectively.
So, by definition, $$\textrm{sn}[K(k)-\tau;k]=\sin{(\arcsin z)}=z.$$
Then, since $z=\frac{\sin{\frac{\theta}{2}}}{\sqrt{k}}$, $$k\,\textrm{sn}^2[K(k)-\tau;k]=k\,z^2=\sin^2{\frac{\theta}{2}}.$$