Problem $\S12$ $2(b)$ of Landau & Lifshitz "Mechanics" Integration of the Equations of motion [closed]

Question

I'm studying Landau, Lifshitz - Mechanics. Could someone help me with this problem ? =)

Problem $\S12$ $2(b)$ (Page 27 3rd Edition) Determine the period of oscillation, as a function of the energy, when a particle of mass $m$ moves in fields for which the potential energy is

$(b) \quad U=-U_{0}/\cosh^{2}\alpha x \quad -U_0<E<0.$

Attempt at solution:

The period is given by

$$ T=4 \sqrt{\frac{m}{2}}\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}.$$

How can I evaluate this integral? I know that the answer is $$T=(\pi/\alpha)\sqrt{2m/|E|}.$$

Answer 1

Hints:

1. From energy conservation $$\frac{1}{2}m\dot{x}^2 ~=~ E-U,$$ with even/symmetric potential $$U ~=~ - \frac{U_0}{c^2},\qquad c~:=~\cosh(\alpha x), \qquad 0<-E<U_0,$$ one gets the quarter period $$\frac{T}{4}~=~ \sqrt{\frac{m}{2}} \int_0^{x_1}\frac{dx}{\sqrt{E-U}},$$ where $x_1>0$ is the upper turning point determined by the condition $E=U(x_1)$.

2. Prove that $$\int_0^{x_1}\frac{dx}{\sqrt{E-U}}~=~\frac{1}{\alpha\sqrt{|E|}}\int_0^{\sqrt{a}}\frac{ds}{\sqrt{(a-s^2)}}, $$ where $$ s~:=~\sinh(\alpha x),\qquad ~a:=~ -U_0/E-1~>~0.$$

3. Show that the last integral does not depend on $a>0$. Choose $a=1$, and perform a substitution $s=\sin(t)$.

Answer 2

Some idea to evaluate the integral: $$\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}=\int \cosh\alpha x\frac{dx}{\sqrt{E \cosh^{2}\alpha x+U_{0}}} = \int \frac{d(\sinh\alpha x)/\alpha}{\sqrt{(E+U_0) + E \sinh^{2}\alpha x }}$$ The last equality use the identity $\cosh^2 x - \sinh^2 x = 1$.

Using $y=\sinh \alpha x$ and factoring $|E|$ out of the square root (because $E<0$), you should then be able see it as a standard integral of the form $\int dy/\sqrt{a^2-y^2} = \sin^{-1}(y/a)$