How I Impressed My Wife: Part 5i

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:

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

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

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

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

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

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

if $|b| < 1$ . (The cases $|b| = 1$ and $|b| > 1$ have already been handled earlier in this series.)

I will evaluate this integral using partial fractions. The denominator factors as the product of two irreducible quadratics, and so I must solve

$\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{Cu + D}{[u + \sqrt{1-b^2}]^2 +b^2}$ ,

$\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{u^2 - 2u \sqrt{1-b^2} +1} + \frac{Cu + D}{u^2 + 2u\sqrt{1-b^2} + 1}$ .

I now clear out the denominators:

$2(1+u^2) = (Au+B)(u^2 + 2u\sqrt{1-b^2} + 1) + (Cu+D)(u^2 - 2u\sqrt{1-b^2} + 1)$

Now I multiply out the right-hand side:

$Au^3 + 2Au^2 \sqrt{1-b^2} + Au + Bu^2 + 2Bu\sqrt{1-b^2} + B$

$+ Cu^3 - 2Cu^2 \sqrt{1-b^2} + Cu + Du^2 - 2Du\sqrt{1-b^2} + D$

Equating this with $2u^2 + 2$ and matching coefficients yields the following system of four equations in four unknowns:

$A + C = 0$

$2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$

$Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$

$B + D = 2$

Ordinarily, four-by-four systems of equations are rather painful to solve, but this system isn’t so bad.

From the first equation, I see that $C = -A$ .

From the third equation, I see that

$Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$

$Au + 2Bu \sqrt{1-b^2} -Au - 2Du \sqrt{1-b^2} = 0$

$2Bu \sqrt{1-b^2} - 2Du \sqrt{1-b^2} = 0$

$B - D = 0$

$B = D$ .

From the fourth equation, I see that

$B + D = 2$

$B + B = 2$

$2B = 2$

$B = 1$ ,

so that $D =1$ as well. Finally, from the second equation, I see that

$2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$

$2A\sqrt{1-b^2} + 1 -2(-A) \sqrt{1-b^2} + 1 = 2$

$4A\sqrt{1-b^2} = 0$

$A= 0$ ,

so that $C = 0$ as well. This yields the partial fractions decomposition

$\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2}$ .

This can be confirmed by directly adding the fractions on the right-hand side:

$\displaystyle \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2}$

$= \displaystyle \frac{[u + \sqrt{1-b^2}]^2 +b^2 + [u - \sqrt{1-b^2}]^2 +b^2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

$= \displaystyle \frac{u^2 + 2u\sqrt{1-b^2} + 1 - b^2 + b^2 + u^2 - 2u\sqrt{1-b^2} + 1 - b^2 + b^2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

$= \displaystyle \frac{2u^2 + 2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$ .

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$ , in tomorrow’s post.

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How I Impressed My Wife: Part 5i

Published by John Quintanilla

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