# How I Impressed My Wife: Part 5c

Earlier in this series, I gave three different methods of showing that

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

Using the fact that $Q$ is independent of $a$, I’ll now give a fourth method.
Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. For example, let me substitute $a =0$:

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

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

So that I can employ the magic substitution $u = \tan x/2$, I’ll divide the interval of integration into two pieces and then perform the substitution $x = t + 2\pi$ on the second piece:

$Q = \displaystyle \int_{0}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x} + \int_{\pi}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{0}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x} + \int_{-\pi}^{0} \frac{dt}{\cos^2 (t+2\pi) + b^2 \sin^2 (t+2\pi)}$

$= \displaystyle \int_{0}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x} + \int_{-\pi}^{0} \frac{dt}{\cos^2 t + b^2 \sin^2 t}$

$= \displaystyle \int_{0}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x} + \int_{-\pi}^{0} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

I’ll continue with this fourth evaluation of the integral in tomorrow’s post.