How I Impressed My Wife: Part 2a

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

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

In this series, I’ll explore different ways of evaluating this integral.

green line

I begin by adjusting the range of integration:

$Q = Q_1 + Q_2 + Q_3$ ,

where

$Q_1 = \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ ,

$Q_3 = \displaystyle \int_{3\pi/2}^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

I’ll begin with $Q_3$ and apply the substitution $u = x - 2\pi$ , or $x = u + 2\pi$ . Then $du = dx$ , and the endpoints change from $3\pi/2 \le x 2\pi$ to $-\pi/2 \le u \le 0$ . Therefore,

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 (u+2\pi) + 2 a \sin (u+2\pi) \cos (u+2\pi) + (a^2 + b^2) \sin^2 (u+2\pi)}$ .

Next, we use the periodic property for both sine and cosine — $\sin(x + 2\pi) = \sin x$ and $\cos(x + 2\pi) = \cos x$ — to rewrite $Q_3$ as

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$ .

Changing the dummy variable from $u$ back to $x$ , we have

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ .

Therefore, we can combined $Q_3 + Q_1$ into a single integral:

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

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

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Next, we work on the middle integral $Q_2$ . We use the substitution $u = x - \pi$ , or $x = u + \pi$ , so that $du = dx$ . Then the interval of integration changes from $\pi/2 \le x \le 3\pi/2$ to $-\pi/2 \le u \le \pi/2$ , so that

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 (u+\pi) + 2 a \sin (u+\pi) \cos (u+\pi) + (a^2 + b^2) \sin^2 (u+\pi)}$ .

Next, we use the trigonometric identities

$\sin(u + \pi) = \sin u \cos \pi + \cos u \sin \pi = \sin u \cdot (-1) + \cos u \cdot 0 = - \sin u$ ,

$\cos(u + \pi) = \cos u \cos \pi - \sin u \sin \pi = \cos u \cdot (-1) - \sin u \cdot 0 = - \cos u$ ,

so that the last integral becomes

$Q_2 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{(-\cos u)^2 + 2 a (-\sin u)(- \cos u) + (a^2 + b^2) (-\sin u)^2}$

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

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

On the line above, I again replaced the dummy variable of integration from $u$ to $x$ . We see that $Q_2 = Q_1 + Q_3$ , and so

$Q = Q_1 + Q_2 + Q_3$

$Q = 2 Q_2$

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

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

by John Quintanilla on July 20, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged cosine, integral, periodic trig identities, sine, sum and difference trig identities, trigonometry

Posted by John Quintanilla on July 20, 2015

https://meangreenmath.com/2015/07/20/how-i-impressed-my-wife-part-2a/

Leave a comment

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 )

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 (147)
- Algebra II (239)
- Bases (43)
- Calculus (220)
- Chemistry (9)
- Computer science (9)
- Discrete mathematics (183)
- Elementary (143)
- Engagement (579)
- Geometry (186)
- Guest presenter (330)
- History (50)
- Humor (212)
- News Clips (104)
- Physics (34)
- Popular Culture (182)
- Pre-Algebra (133)
- Precalculus (453)
- Preparing for college (12)
- Probability (76)
- Statistics (75)
- Tales from the Classroom (194)
- Technology (74)
- Theorems (148)
- Uncategorized (49)
Tags
angle antilogarithm arccosine arcsine arctangent area arithmetic series 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 popular culture primes PRIMUS proof Pythagorean theorem Pythagorean trig identities quadratic formula residue sequence series sine sphere sports square root system of equations tangent Taylor series textbook problems triangle trigonometry unsolved problems volume xkcd
Meta

%d bloggers like this: