How I Impressed My Wife: Part 5h

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.

green line

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:

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

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

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

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

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

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$ . In today’s post, I’ll start considering the case $|b| < 1$ .

Factoring the denominator is a bit more complicated if $|b| < 1$ . Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$ , the right-hand side is now a complex number. So, To solve for $u$ , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$ .

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$ , or $|u| = 1$ . This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$ .

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$ ,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$ ,

$\sin 2 \phi = 2 \sin\phi \cos \phi$ ,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$ ,

or

$u = \pm( \sqrt{1-b^2} + i |b|)$ .

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$ .

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

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

To double-check my work, I can directly multiply this product:

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

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

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

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

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$ .

So, at last, I can rewrite the integral $Q$ as

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

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

by John Quintanilla on August 20, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged complex numbers, cosine, De Moivre's Theorem, double angle trig identities, integral, polar coordinates, sine, square root, trigonometry

Posted by John Quintanilla on August 20, 2015

https://meangreenmath.com/2015/08/20/how-i-impressed-my-wife-part-5h/

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