# 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:

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

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

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

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

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

$=\displaystyle \left[ 2\tan^{-1} x \right]^{\infty}_{-\infty}$

$=\displaystyle \left[ 2 \frac{\pi}{2} - 2 \frac{-\pi}{2} \right]$

$= 2\pi$

This matches the expected answer of $Q = \displaystyle \frac{2\pi}{|b|}$ since I used the assumption that $|b| = 1$.

I’ll continue with this fourth evaluation of the integral, examining the two remaining cases, in future posts.

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