How I Impressed My Wife: Part 6d

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.

nvenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

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

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

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

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

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

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

$= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]$

where I’ve assumed $|b| > 1$ , the contour $C_R$ in the complex plane is shown below (graphic courtesy of Mathworld),

and the positive constants $r_1$ and $r_2$ are given by

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

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

Now we have the small matter of simplifying our expression for $Q$ . Actually, this isn’t a small matter because Mathematica 10.1 is not able to simplify this expression much at all:

Fortunately, humans can still do some things that computers can’t. As observed yesterday, The numbers $r_1$ and $r_2$ are chosen so that $\pm ir_1$ and $\pm ir_2$ are the roots of the denominator $z^4 + (4 b^2 - 2) z^2 + 1$ , so that