How I Impressed My Wife: Part 5e

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

To evaluate this integral, I need to find the four complex roots of the denominator:

$u^4 + (4b^2 - 2) u^2 + 1 = 0$

$u^2 = \displaystyle \frac{ 2 - 4b^2 \pm \sqrt{(4b^2 - 2)^2 - 4}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2 - 4 + 4}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}$

$u^2 = \displaystyle \frac{2 - 4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}$

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

To solve for $u$ , there are three separate cases that have to be considered: $|b| = 1$ , $|b| > 1$ , and $|b| < 1$ . I’ll begin with the easiest case of $|b| = 1$ . In this case, the integral $Q$ is easy to evaluate: