Integral $\int_0^{2\pi} \sqrt{A^2-\frac{B^2}{\sin^2x}} dx$

Question

This is the integral obtained when finding the energy level of a hydrogen atom by Sommerfeld quantization. How to evaluate this integral? $$\int_0^{2\pi}\sqrt{A^2-\frac{B^2}{\sin^2\theta}}d\theta$$ Thank you.

Answer 1

Computing the antiderivative$$I=\int\sqrt{A^2-\frac{B^2}{\sin^2(x)}}\,dx$$ is not so bad.

Let $$x=\cot ^{-1}(t) \qquad \text{and} \qquad C^2=\frac {A^2-B^2}{B^2}$$

$$I=-B\int \frac{\sqrt{C^2-t ^2}}{t^2+1}\,dt$$ $$t=C\sin(u)\qquad \implies \qquad I=-B C^2 \int \frac{\cos ^2(u)}{1+C^2 \sin ^2(u)}\,du$$ $$u=\tan^{-1}(v)\qquad \implies \qquad I=-B C^2 \int \frac{dv}{\left(v^2+1\right) \left(\left(C^2+1\right) v^2+1\right)}$$ Now, partial fraction decomposition would lead to two arctangents.