How I Impressed My Wife: Part 3b

Previously in this series, I showed that

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

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

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

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

We now employ the substitution $\theta = 2x$ , so that $d\theta = 2 \, dx$ . Also, the limits of integration change from $0 \le x \le 2\pi$ to $0 \le \theta \le 4\pi$ , so that

$Q = \displaystyle \int_0^{4\pi} \frac{d\theta}{1+\cos \theta + 2 a \sin \theta + (a^2 + b^2)(1-\cos \theta)}$

$= \displaystyle \int_0^{4\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

Next, I’ll divide write $Q = Q_1 + Q_2$ by dividing the interval of integration (not to be confused with the $Q_1$ and $Q_2$ used in the previous method), where

$Q_1 = \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$Q_2 = \displaystyle \int_{2\pi}^{4\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

For $Q_2$ , I employ the substitution $u = \theta - 2\pi$ , so that $\theta = u + 2\pi$ and $du = d\theta$ . Under this substitution, the interval of integration changes from $2\pi \le \theta \le 4\pi$ to $0 \le u \le 2\pi$, and so

$Q_2 = \displaystyle \int_{0}^{2\pi} \frac{du}{(1+a^2+b^2) + 2 a \sin (u+2\pi) + (1 - a^2 - b^2) \cos (u+2\pi)}$

Next, I use the periodic property for both sine and cosine — $\sin(u + 2\pi) = \sin u$ and $\cos(u+ 2\pi) = \cos u$ — to rewrite $Q_2$ as

$Q_2 = \displaystyle \int_{0}^{2\pi} \frac{du}{(1+a^2+b^2) + 2 a \sin u + (1 - a^2 - b^2) \cos u}$

Except for the dummy variable $u$ , instead of $\theta$ , we see that $Q_2$ is identical to $Q_1$ . Therefore,

$Q = Q_1 + Q_2 = 2 Q_1 = 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$ .

I’ll continue this different method of evaluating this integral in tomorrow’s post.

How I Impressed My Wife: Part 3a

Previously in this series, I showed that

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

For this next technique, I begin by using the trigonometric identities

$\sin^2 x = \displaystyle \frac{1-\cos 2x}{2}$ ,

$\cos^2x = \displaystyle \frac{1+\cos 2x}{2}$ ,

$2 \sin x \cos x = \sin 2x$ .

Using these identities, we obtain

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\displaystyle \frac{1+\cos 2x}{2} + a \sin 2x + (a^2 + b^2) \displaystyle \frac{1-\cos 2x}{2}}$

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

In this way, the exponents have been removed from the denominator, thus making the integrand somewhat less complicated.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

How I Impressed My Wife: Part 2f

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.So far in this series, I’ve shown that

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

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

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

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}$

This last integral can be evaluated using a standard trick. Let $\theta = \tan^{-1} w$ , so that $w = \tan \theta$ . We differentiate this last equation with respect to $w$ :

$\displaystyle \frac{dw}{dw} = \sec^2 \theta \cdot \displaystyle \frac{d\theta}{dw}$

Employing a Pythagorean identity, we have

$1 = (1+ \tan^2 \theta) \cdot \displaystyle \frac{d\theta}{dw}$

Since $w = \tan \theta$ , we may rewrite this as

$1 = (1+ w^2) \cdot \displaystyle \frac{d\theta}{dw}$

$\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d\theta}{dw}$

$\displaystyle \frac{1}{1+w^2} = \displaystyle \frac{d}{dw} \tan^{-1} w$

Integrating both sides with respect to $w$ , we obtain the antiderivative

$\displaystyle \int \frac{1}{1+w^2} = \tan^{-1} w + C$

We now employ this antiderivative to evaluate $Q$ :

$Q = \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{ dw }{w^2 +1}$

$= \displaystyle \frac{2}{|b|} \displaystyle \left[ \tan^{-1} w \right]^{\infty}_{-\infty}$

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

$= \displaystyle \frac{2\pi}{|b|}$

And so, at long last, we have arrived at the solution for the integral $Q$ . Surprisingly, the answer is independent of the parameter $a$ .

These last few posts illustrated the technique that I used to compute this integral for my wife in support of her recent paper in Physical Review A. However, I had more than a few false starts along the way… or, at the time, I thought they were false starts. It turns out that there are multiple ways of evaluating this integral, and I’ll explore another method of attack beginning with tomorrow’s post.

How I Impressed My Wife: Part 2e