# How I Impressed My Wife: Part 3i

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

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

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

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

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= 2 \displaystyle \int_{-\pi}^{\pi} \frac{d\phi}{S + R \cos \phi}$

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

$= \displaystyle \frac{2}{|b|} \int_{-\infty}^{\infty} \frac{dv}{1 + v^2}$,

as long as $b \ne 0$. (In the above calculations, the constants $R$, $S$, and $\alpha$ depend on $a$ and $b$ but are no longer necessary at this point in the calculation.)

We can now directly compute this final integral using an antiderivative derived earlier in this series:

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

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

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

And so, with this completely different technique, we arrive at the same answer for the integral $Q$.

However, there are still different ways of computing $Q$. I’ll start on another method of attack with tomorrow’s post.