Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$ Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?

Here is my attempt. Divide the first equation by $\dot\phi$, then we get
$$ \frac{d\ln\dot\phi}{dt}+2\frac{d\ln(\sin\theta)}{dt}=0.$$
We then obtain the integral of motion
$$\ln\dot\phi+2\ln\sin\theta=C.$$
Hence, we find $\dot\phi\sin^2\theta=e^C=B$.
Plugging this in the second equation, we find
$$ \ddot\theta-B^2\frac{\cos\theta}{\sin^3\theta}\dot\theta=0.$$
Multiplying this by $\dot\theta$, we find
$$\frac12\frac{d\dot\theta^2}{dt}+\frac b2\frac{d\sin^{-2}\theta}{dt}=0,$$
where $b=B^2$.
Then we get another integral of motion, a Hamiltonian $H$.
$$\frac{\dot\theta^2}2+\frac b{2\sin^2\theta}=H.$$
I believe this is right, but I don't have time to check it carefully now.
I hope this helps.
