# How I Impressed My Wife: Part 5a

Amazingly, the integral below has a simple solution: $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$. For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$. Earlier in this series, I showed that $Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ $= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$ $= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$ $= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

I now multiply the top and bottom of this last integral by $a^2 + b^2$: $Q = \displaystyle \frac{2}{(a^2+b^2)^2} \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$ $= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

I now employ the substitution $w = (a^2 + b^2) v$, so that $dw = (a^2 + b^2) v$. Since $a^2 + b^2 > 0$, the endpoints of integration do not change, and so $Q = \displaystyle 2 \int_{-\infty}^{\infty} \frac{dw}{w^2 + b^2 }$.

This final integral is independent of $a$. Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I wish. For example, I can let $a = 0$ without altering the value of $Q$: $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x} = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

This provides a considerable simplification of the integral $Q$ which also opens up additional methods of evaluation.

## One thought on “How I Impressed My Wife: Part 5a”

This site uses Akismet to reduce spam. Learn how your comment data is processed.