Show that \[ \int_{0}^{2 \pi} \frac{d \theta}{(a+\cos \theta)^{2}}=\frac{2 \pi a}{\left(a^{2}-1\right)^{3 / 2}}, \quad \text { whenever } a>1 .\]
Solution
First evaluate \[ J(a)=\int_0^{2\pi}\frac{d\theta}{a+\cos\theta}, \qquad a>1. \] With \(z=e^{i\theta}\), \[ \cos\theta=\frac12\left(z+\frac1z\right), \qquad d\theta=\frac{dz}{iz}, \] so \[ J(a) =\frac2i\int_{|z|=1} \frac{dz}{z^2+2az+1}. \] The roots of the denominator are \[ z_\pm=-a\pm\sqrt{a^2-1}. \] For \(a>1\), one has \[ |z_+|<1<|z_-|, \] so only \(z_+\) lies inside the unit circle. Its residue is \[ \operatorname{Res}_{z=z_+} \frac1{z^2+2az+1} =\frac1{2(z_++a)} =\frac1{2\sqrt{a^2-1}}. \] Hence \[ J(a) =\frac2i\cdot2\pi i\cdot \frac1{2\sqrt{a^2-1}} =\frac{2\pi}{\sqrt{a^2-1}}. \]
Since \(a>1\), the denominator \(a+\cos\theta\) is bounded away from zero, so differentiation under the integral sign is valid. Therefore \[ J'(a) =-\int_0^{2\pi} \frac{d\theta}{(a+\cos\theta)^2}. \] Differentiating the explicit expression for \(J\) gives \[ J'(a) =-\frac{2\pi a}{(a^2-1)^{3/2}}. \] Thus \[ \boxed{ \int_0^{2\pi} \frac{d\theta}{(a+\cos\theta)^2} =\frac{2\pi a}{(a^2-1)^{3/2}}.} \]