单摆（大摆角）

\begin{equation} E = \frac{1}{2} m l^2 \dot \theta^2 - mg l \cos\theta = - mg l \cos\theta_0~, \end{equation}

\begin{equation} \dot{\theta} = \sqrt{\frac{2g}{l} (\cos\theta - \cos\theta_0) }~. \end{equation}

\begin{equation} t(\alpha) = \sqrt{\frac{l}{2g}} \int_0^{\alpha} \frac{ \,\mathrm{d}{\theta} }{\sqrt{\cos\theta - \cos\theta_0}} = \sqrt{\frac{l}{g}} \csc\frac{\theta_0}{2} F \left(\frac{\alpha}{2}, \csc\frac{\theta_0}{2} \right) ~, \qquad (0 \leqslant \alpha \leqslant \theta_0)~. \end{equation}

\begin{equation} T = 4t(\theta_0) = 4 \sqrt{\frac{l}{g}} \csc\frac{\theta_0}{2} F \left(\frac{\theta_0}{2}, \csc\frac{\theta_0}{2} \right) ~. \end{equation}

\begin{equation} T = 4 \sqrt{\frac{l}{g}} F \left(\frac{\pi}{2}, \sin\frac{\theta_0}{2} \right) ~. \end{equation}

\begin{equation} t(\alpha)=\frac12\sqrt{\frac{l}{g}}\int_{0}^{\alpha}\frac{\mathrm{d}\theta}{\sqrt{\sin^2\frac{\theta_0}2-\sin^2\frac{\theta}2}}~. \end{equation}

\begin{equation} \sin \phi(\theta)=\frac{\sin^2\frac{\theta}2}{\sin^2\frac{\theta_0}2}~, \end{equation}

\begin{equation} \mathrm d\theta=2\frac{\sin\frac{\theta_0}2}{\cos\frac{\theta}2}\cos\phi(\theta)\mathrm d\phi =2\frac{\sin\frac{\theta_0}2}{\sqrt{1-\sin^2\frac{\theta}2}}\sqrt{1-\sin^2\phi(\theta)}\mathrm d\phi =2\frac{\sqrt{\sin^2\frac{\theta_0}2-\sin^2\frac{\theta}2}}{\sqrt{1-\sin^2\frac{\theta_0}2\sin^2{\phi(\theta)}}}\mathrm d\phi~. \end{equation}

\begin{equation} t(\alpha)=\sqrt{\frac{l}{g}}\int_{0}^{\phi(\alpha)}\frac{\mathrm{d}\phi}{\sqrt{1-\sin^2\frac{\theta_0}2\sin^2{\phi(\theta)}}}~. \end{equation}

\begin{equation} T=4t(\theta_0)=\sqrt{\frac{l}{g}}\int_{0}^{\frac\pi2}\frac{\mathrm{d}\phi}{\sqrt{1-\sin^2\frac{\theta_0}2\sin^2{\phi(\theta)}}}=4 \sqrt{\frac{l}{g}} F \left(\frac{\pi}{2}, \sin\frac{\theta_0}{2} \right) ~, \end{equation}