Finding period from action-angle variable in one dimensional potential I want to calculate the period from the action-angle variable for a particle in a one dimensional potential $V = V_0 \tan^2(q \pi/2a)$. 
After doing some algebra I get $$I = \frac{\sqrt{2mE}}{2\pi}  \int_{q_1}^{q_2}\sqrt{1-\frac{V_0}{E}\tan^2{\frac{q\pi}{2a}}}dq.$$
Solving which gives me $$I = \frac{a}{\pi}\sqrt{2mE}\left(\sqrt{1+ \frac{V_0}{E}} - \sqrt{\frac{V_0}{E}}\right).$$ How can I get the period of the motion and how will it behave when $E\gg V_0$ and vice versa? Also how does this picture change if I change $a$ adiabatically with time?
 A: The trick is to re-express the Hamiltonian in terms of the action variables. In your case this simply means to invert the expression $I(E)$ into an expression $E(I) \equiv H(I)$. This is then your Hamiltonian expressed in terms of action-angle coordinates. Finally, you can compute the evolution of the angle $\phi$ canonically conjugate to $I$ from Hamilton's equations of motion
$$\dot{\phi} = \frac{\partial H}{\partial I}$$
Notice that $\dot{I} = -\partial H/\partial \phi = 0$, $I$ is constant, and thus also the frequency $\dot{\phi}$ is constant for any trajectory. 
You are also asking about the behaviour when $E \ll V_0$ as opposed to $E \gg V_0$. This is respectively equivalent to $I \ll a\sqrt{m E}$ and $I \gg a\sqrt{m E}$, as  you should directly verify. You will then see that in the leading-order limit of a small action, the rate $\dot{\phi}$ is independent of $I$ - the system behaves like a harmonic oscillator. However, for large actions (large energies) the frequency will have a steep dependence on $I$.
Good luck!
