How I Impressed My Wife: Part 6g

This series was inspired by a question that my wife asked me: calculate

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

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.

Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

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:

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

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

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

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

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

$= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$

$= 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right]$ ,

where I’ve made the assumption that $|b| < 1$ . In the above derivation, $C_R$ is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Also,

$r_1 = \sqrt{1-b^2} + |b|i$

and

$r_2 = -\sqrt{1-b^2} + |b|i$

are the two poles of the final integrand that lie within this contour.

It now remains to simplify the final algebraic expression. To begin, I note

$\displaystyle \frac{r_1}{r_1^2-1} = \displaystyle \frac{\sqrt{1-b^2} + |b|i}{[\sqrt{1-b^2} + |b|i]^2 - 1}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{1-b^2 + 2|b|i\sqrt{1-b^2} - |b|^2 - 1}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{-2|b|^2 + 2|b|i\sqrt{1-b^2}}$

$= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{2|b|i(|b|i +\sqrt{1-b^2})}$

$= \displaystyle \frac{1}{2|b|i}$ .

Similarly,

$\displaystyle \frac{r_2}{r_2^2-1} = \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{[-\sqrt{1-b^2} + |b|i]^2 - 1}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{1-b^2 - 2|b|i\sqrt{1-b^2} - |b|^2 - 1}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{-2|b|^2 - 2|b|i\sqrt{1-b^2}}$

$= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{2|b|i(|b|i -\sqrt{1-b^2})}$

$= \displaystyle \frac{1}{2|b|i}$ .

Therefore,

$Q = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right] = 2\pi i \left[ \displaystyle \frac{1}{2|b|i} + \frac{1}{2|b| i} \right] = 2\pi i \displaystyle \frac{2}{2|b|i} = \displaystyle \frac{2\pi}{|b|}$ .

And so, at long last, I’ve completed a fifth different evaluation of $Q$ .

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How I Impressed My Wife: Part 6g

Published by John Quintanilla

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