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 lineI 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}

green line

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

1 Comment
by John Quintanilla on July 20, 2015  •  Permalink
Posted in Calculus, Precalculus
Tagged , , , , ,

Posted by John Quintanilla on July 20, 2015

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

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  1. How I Impressed My Wife: Index | Mean Green Math

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