How I Impressed My Wife: Part 5d

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 $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

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

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

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

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

I now employ the magic substitution $u = \tan x/2$ , so that

$\cos x = \displaystyle \frac{1-u^2}{1+u^2}$ ,

$\sin x = \displaystyle \frac{2u}{1+u^2}$ ,

$dx = \displaystyle \frac{2 \, du}{1+u^2}$ .

The endpoints change from $-\pi < x < \pi$ to $-\infty < u < \infty$ , and so

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ \displaystyle \frac{2}{1+u^2} du}{ \left[\displaystyle \frac{1-u^2}{1+u^2} \right]^2 + b^2 \left[\displaystyle \frac{2u}{1+u^2}\right]^2}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ \left[1-u^2 \right]^2 + b^2 \left[2u \right]^2}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{1-2u^2+u^4 + 4 b^2 u^2}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

I have transformed the integral $Q$ into a new integral involving a fairly simple rational function that can be evaluated using standard (and non-standard) techniques.

Hypothetically, the magic substitution can be applied to the original integral. Unfortunately, I was unable to make any headway in finding the four complex roots of the resulting rational function. However, since I made the replacement $a =0$ at the start, this new rational function is much more tractable.

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

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How I Impressed My Wife: Part 5d

Published by John Quintanilla

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Published by John Quintanilla

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