# How I Impressed My Wife: Part 2d

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 $Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ $= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$ $= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$

To evaluate this last integral, I complete the square in the denominator. I first factor $(a^2+b^2)$ out of the denominator: $Q = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{u^2 + \displaystyle \frac{2 a}{a^2+b^2} u + \displaystyle \frac{1}{a^2+b^2} }$ $= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u^2 + \displaystyle \frac{2 a}{a^2+b^2} u + \displaystyle \frac{a^2}{(a^2+b^2)^2} \right) + \displaystyle \frac{1}{a^2+b^2} - \displaystyle \frac{a^2}{(a^2+b^2)^2} }$ $= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{a^2+b^2}{(a^2+b^2)^2} - \displaystyle \frac{a^2}{(a^2+b^2)^2} }$ $= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{du}{\left( u + \displaystyle \frac{a}{a^2+b^2} \right)^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

Next, I employ the substitution $v = u + \displaystyle \frac{a}{a^2+b^2}$, so that $dv = du$. The endpoint of the integral do not change with this substitution, and so $Q = \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$ I’ll continue with the evaluation of this integral in tomorrow’s post.