How I Impressed My Wife: Part 5b

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

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

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

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

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

Yesterday, I showed used the substitution $w = (a^2 + b^2) v$ to show that $Q$ was independent of $a$ . Today, I’ll use a different method to establish the same result. Let

$Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }$ .

Notice that I’ve written this integral as a function of the parameter $a$ . I will demonstrate that $Q'(a) = 0$ , so that $Q(c)$ is a constant with respect to $a$ . In other words, $Q(a)$ does not depend on $a$ .

To do this, I differentiate under the integral sign with respect to $a$ (as opposed to $x$ ) using the Quotient Rule:

$Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

I now apply the trigonometric substitution $v = \displaystyle \frac{b}{a^2+b^2} \tan \theta$ , so that

$(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta$

and

$dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

The endpoints of integration change from $-\infty < v < \infty$ to $-\pi/2 < \theta < \pi/2$ , and so

$Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta$

$= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]$

$= 0$ .

I’m not completely thrilled with this demonstration that $Q$ is independent of $a$ , mostly because I had to do so much simplification of the integral $Q$ to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

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

and demonstrate that $Q$ is independent of $a$ , perhaps by differentiating $Q$ with respect to $a$ and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying $Q$ first.

1 Comment

by John Quintanilla on August 13, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged derivative, double angle trig identities, integral, Pythagorean trig identities, ratio trig identities, reciprocal trig identities, secant, tangent

Posted by John Quintanilla on August 13, 2015

https://meangreenmath.com/2015/08/13/how-i-impressed-my-wife-part-5b-2/

1 Comment

How I Impressed My Wife: Index | Mean Green Math

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

You are commenting using your Google+ account. ( Log Out / Change )

Cancel

Connecting to %s

Search for:
Follow Blog via Email

Enter your email address to follow this blog and receive notifications of new posts by email.
Top Posts & Pages
Archives
Categories
- Algebra I (146)
- Algebra II (231)
- Bases (43)
- Calculus (220)
- Chemistry (8)
- Computer science (9)
- Discrete mathematics (151)
- Elementary (140)
- Engagement (540)
- Geometry (172)
- Guest presenter (300)
- History (48)
- Humor (204)
- News Clips (98)
- Physics (32)
- Popular Culture (151)
- Pre-Algebra (133)
- Precalculus (430)
- Preparing for college (12)
- Probability (68)
- Statistics (75)
- Tales from the Classroom (194)
- Technology (74)
- Theorems (143)
- Uncategorized (48)
Tags
2048 angle antilogarithm arccosine arcsine arctangent area bag of tricks binary circle combinatorics Common Core complex numbers compound interest conceptual barrier confidence interval convex polygons cosine decimal De Moivre's Theorem derivative differential equation divisibility division algorithm e exponent exponential factorial factoring fraction function gamma geometric series graph high-stakes testing hypothesis test index to a series of posts induction integral inverse function law of cosines law of sines limit logarithm logic MAA magic trick multiplication partial fractions Pascal's triangle polar coordinates polynomial primes PRIMUS proof Pythagorean theorem Pythagorean trig identities quadratic formula regression residue sequence series sine sphere sports square root system of equations tangent Taylor series textbook problems triangle trigonometry unsolved problems volume xkcd
Meta

Mean Green Math

How I Impressed My Wife: Part 5b

Share this:

Like this:

Related

1 Comment

Leave a Reply Cancel reply

Follow Blog via Email

Top Posts & Pages

Archives

Categories

Tags

Meta