How I Impressed My Wife: Part 2a

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

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

In this series, I’ll explore different ways of evaluating this integral.

I begin by adjusting the range of integration:

$Q = Q_1 + Q_2 + Q_3$ ,

where

$Q_1 = \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_3 = \displaystyle \int_{3\pi/2}^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

I’ll begin with $Q_3$ and apply the substitution $u = x - 2\pi$ , or $x = u + 2\pi$ . Then $du = dx$ , and the endpoints change from $3\pi/2 \le x 2\pi$ to $-\pi/2 \le u \le 0$ . Therefore,

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 (u+2\pi) + 2 a \sin (u+2\pi) \cos (u+2\pi) + (a^2 + b^2) \sin^2 (u+2\pi)}$ .

Next, we use the periodic property for both sine and cosine — $\sin(x + 2\pi) = \sin x$ and $\cos(x + 2\pi) = \cos x$ — to rewrite $Q_3$ as

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$ .

Changing the dummy variable from $u$ back to $x$ , we have

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

Therefore, we can combined $Q_3 + Q_1$ into a single integral:

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

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

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Next, we work on the middle integral $Q_2$ . We use the substitution $u = x - \pi$ , or $x = u + \pi$ , so that $du = dx$ . Then the interval of integration changes from $\pi/2 \le x \le 3\pi/2$ to $-\pi/2 \le u \le \pi/2$ , so that