Integral (4)

Problem:

$ \int\frac{d\theta}{A + B \cos\theta} $

Solution:

Let $ t = \tan\frac{\theta}{2} $, $ \cos \theta = \frac{1 - t^2}{1 + t^2} $, $ d\theta = \frac{2}{1 + t^2}dt $

$ \begin{eqnarray*} & & \int\frac{d\theta}{A + B \cos\theta} \\ &=& \int\frac{\frac{2}{1 + t^2}dt}{A + B \frac{1 - t^2}{1 + t^2}} \\ &=& \int\frac{2dt}{A(1 + t^2) + B(1 - t^2)} \\ &=& \int\frac{2dt}{(A + B) + (A - B)t^2} \\ &=& 2\sqrt{\frac{1}{(A + B)(A - B)}}\tan^{-1}\left(\sqrt{\frac{A - B}{A + B}}t\right) \\ &=& 2\sqrt{\frac{1}{(A + B)(A - B)}}\tan^{-1}\left(\sqrt{\frac{A - B}{A + B}}\tan\frac{x}{2}\right) \\ \end{eqnarray*} $

The second last step comes from a previously derived result which can be found here.