# 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.

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