Question

what is the answer for \int^{2\pi}_{0} \frac{d\phi}{1-A\cos(\phi)}

Answer 1

\[\int\limits^{2\pi}_{0} \frac{d\phi}{1-A\cos(\phi)}\]

Answer 2

yeah, do you know how to calculate it? I couldn't and I didn't find it in any integral list

Answer 3

\[\int\limits \frac{d \phi}{1 - A \cos \phi}\]\[\tan \frac{1}{2} \phi = t\]\[\cos \phi = \frac{1 - t^2}{1 + t^2}\]\[d \phi = \frac{2}{1 + t^2}dt\]\[= \int\limits \frac{\frac{2}{1 + t^2}}{1 - A\left(\frac{1 - t^2}{1 + t^2}\right)}dt\]\[= \int\limits \frac{2}{1 + t^2 - A + At^2}dt\]\[=\int\limits \frac{2}{(1 + A)t^2 + 1 - A}dt\]\[t = \sqrt{\frac{1 - A}{1 + A}}\tan \theta\]\[dt = \sqrt{\frac{1 - A}{1 + A}}\sec^2 \theta d \theta\]\[=\int\limits \frac{2\sqrt{\frac{1 - A}{1 + A}}\sec^2 \theta}{(1 - A)\sec^2 \theta}d \theta\]\[=\int\limits \frac{2\sqrt{\frac{1 - A}{1 + A}}}{1 - A}d \theta\]\[=\frac{2 \sqrt{\frac{1 - A}{1 + A}}}{1 - A}\theta + C\]\[=\frac{2\sqrt{\frac{1 - A}{1 + A}}}{1 - A}\arctan \left(\sqrt{\frac{1 + A}{1 - A}}t \right) + C\]\[=\frac{2\sqrt{\frac{1 - A}{1 + A}}}{1 - A}\arctan \left(\sqrt{\frac{1 + A}{1 - A}}\tan \frac{1}{2} \phi \right) + C\]

Answer 4

Wow! Thank you very much! I really appreciate that!