Show that if \(a,b\in \RR\) with \(a > \abs{b}\), then \[ \int_{0}^{2 \pi} \frac{d \theta}{a+b \cos \theta}=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}} .\]
Solution
Write \(I\) for the integral.
1. If \(b = 0\), then \(I = \dfrac{2\pi}{\sqrt{a^2 - b^2}}\).
The integrand is the constant \(1/a\), so \(I = 2\pi/a = 2\pi/\sqrt{a^2}\).
2. If \(b\neq0\), then\[I = \frac{2}{i} \oint_{\abs{z} = 1} g(z)\,dz, \qquad g(z)\coloneqq\frac{1}{bz^2 + 2az + b},\]with the unit circle oriented counterclockwise.
With \(z = e^{i\theta}\), one has \(d\theta = \frac{dz}{iz}\) and \(\cos \theta = \frac{z + z^{-1}}{2}\), so \[a + b \cos \theta = \frac{bz^2 + 2az + b}{2z} \qquad\text{and}\qquad \frac{d\theta}{a+b\cos\theta} = \frac{2z}{bz^2 + 2az + b}\cdot\frac{dz}{iz}.\]
3. If \(b\neq0\), the only pole of \(g\) inside the unit circle is\[z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}, \qquad\text{and}\qquad \operatorname{Res}(g, z_1) = \frac{1}{2\sqrt{a^2 - b^2}}.\]
The roots of \(bz^2 + 2az + b\) are \(z_1\) and \(z_2 = \frac{-a - \sqrt{a^2 - b^2}}{b}\), and \(z_1 z_2 = 1\). Since \(a > \abs{b} > 0\), \[\abs{z_2} = \frac{a + \sqrt{a^2 - b^2}}{\abs{b}} > \frac{a}{\abs{b}} > 1,\] so \(\abs{z_1} = 1/\abs{z_2} < 1\). The pole at \(z_1\) is simple, with residue \[\frac{1}{b(z_1 - z_2)} = \frac{1}{b \cdot \frac{2\sqrt{a^2 - b^2}}{b}}.\]
4. If \(b\neq0\), then \(I = \dfrac{2\pi}{\sqrt{a^2 - b^2}}\).
By steps 2 and 3 and the residue theorem, \[I = \frac{2}{i} \cdot 2\pi i \operatorname{Res}(g, z_1) = \frac{4\pi}{2\sqrt{a^2 - b^2}}.\]
5. Q.E.D.
Steps 1 and 4 give \(I = \boxed{\dfrac{2\pi}{\sqrt{a^2 - b^2}}}\) in both cases.