How I Impressed My Wife: Part 6g

This series was inspired by a question that my wife asked me: calculate

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

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

green lineSince 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 \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

= 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right],

where I’ve made the assumption that |b| < 1. In the above derivation, C_R is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Also,

r_1 = \sqrt{1-b^2} + |b|i

and

r_2 = -\sqrt{1-b^2} + |b|i

are the two poles of the final integrand that lie within this contour.

It now remains to simplify the final algebraic expression. To begin, I note

\displaystyle \frac{r_1}{r_1^2-1} = \displaystyle \frac{\sqrt{1-b^2} + |b|i}{[\sqrt{1-b^2} + |b|i]^2 - 1}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{1-b^2 + 2|b|i\sqrt{1-b^2} - |b|^2 - 1}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{-2|b|^2 + 2|b|i\sqrt{1-b^2}}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{2|b|i(|b|i +\sqrt{1-b^2})}

= \displaystyle \frac{1}{2|b|i}.

Similarly,

\displaystyle \frac{r_2}{r_2^2-1} = \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{[-\sqrt{1-b^2} + |b|i]^2 - 1}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{1-b^2 - 2|b|i\sqrt{1-b^2} - |b|^2 - 1}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{-2|b|^2 - 2|b|i\sqrt{1-b^2}}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{2|b|i(|b|i -\sqrt{1-b^2})}

= \displaystyle \frac{1}{2|b|i}.

Therefore,

Q = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right] = 2\pi i \left[ \displaystyle \frac{1}{2|b|i} + \frac{1}{2|b| i} \right] = 2\pi i \displaystyle \frac{2}{2|b|i} = \displaystyle \frac{2\pi}{|b|}.

green lineAnd so, at long last, I’ve completed a fifth different evaluation of Q.

1 Comment
by John Quintanilla on August 29, 2015  •  Permalink
Posted in Calculus, Precalculus
Tagged , , , ,

Posted by John Quintanilla on August 29, 2015

https://meangreenmath.com/2015/08/29/how-i-impressed-my-wife-part-6g/

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

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