How I Impressed My Wife: Part 6e

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.

 nvenient 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 \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]

where I’ve assumed |b| > 1, the contour C_R in the complex plane is shown below (graphic courtesy of Mathworld),

and the positive constants r_1 and r_2 are given by

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

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

Now we have the small matter of simplifying our expression for Q. Actually, this isn’t a small matter because Mathematica 10.1 is not able to simplify this expression much at all:

ResidueCalculation

Fortunately, humans can still do some things that computers can’t. The numbers r_1 and r_2 are chosen so that \pm ir_1 and \pm ir_2 are the roots of the denominator z^4 + (4 b^2 - 2) z^2 + 1. In other words,

[ir_1]^4 + (4b^2 - 2) [ir_1]^2 + 1 = r_1^4 - [4b^2-2] r_1^2 + 1 = 0,

r_2^4 - [4b^2-2] r_2^2 + 1 = 0

These relationships will be very handy for simplifying our expression for Q:

Q = 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]

= 2\pi \left[ \displaystyle \frac{r_1^2-1}{2r_1^3-(4b^2-2)r_1} + \frac{r_2^2-1}{2r_2^3-(4b^2-2)r_2} \right]

= 2\pi \left[ \displaystyle \frac{r_1(r_1^2-1)}{2r_1^4-(4b^2-2)r_1^2} + \frac{r_2(r_2^2-1)}{2r_2^4-(4b^2-2)r_2^2} \right]

= 2\pi \left[ \displaystyle \frac{r_1(r_1^2-1)}{r_1^4 + r_1^4-(4b^2-2)r_1^2} + \frac{r_2(r_2^2-1)}{r_2^4+r_2^4-(4b^2-2)r_2^2} \right]

= 2\pi \left[ \displaystyle \frac{r_1(r_1^2-1)}{r_1^4 -1} + \frac{r_2(r_2^2-1)}{r_2^4-1} \right]

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

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

= 2\pi \displaystyle \frac{r_1(r_2^2+1)+(r_1^2+1)r_2}{(r_1^2+1)(r_2^2+1)}

= 2\pi \displaystyle \frac{r_1 r_2^2+r_1+r_1^2 r_2 +r_2}{(r_1^2+1)(r_2^2+1)}

= 2\pi \displaystyle \frac{r_1 +r_2 + r_1 r_2 (r_1 + r_2)}{(r_1^2+1)(r_2^2+1)}

= 2\pi \displaystyle \frac{(r_1 +r_2)(1 + r_1 r_2)}{(r_1^2+1)(r_2^2+1)}

 

To complete the calculation, we recall that

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

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

so that

r_1^2 + r_2^2 = 4b^2 - 2,

r_1^2 r_2^2 = 1,

and hence

r_1 r_2 = 1

since r_1 and r_2 are both positive. Also,

(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2 = 4b^2 -2 + 2 = 4b^2,

so that

r_1 + r_2 = 2|b|.

Finally,

(r_1^2 + 1)(r_2^2 + 1) = (2b^2 + 2|b| \sqrt{b^2-1})(2b^2 - 2|b| \sqrt{b^2 -1})

= 4b^4 - 4b^2 (b^2-1)

= 4b^2.

Therefore,

Q = 2\pi \displaystyle \frac{(r_1 +r_2)(1 + r_1 r_2)}{(r_1^2+1)(r_2^2+1)} = 2\pi \displaystyle \frac{ 2|b| \cdot (1 + 1)}{4b^2} = \displaystyle \frac{8\pi |b|}{4 |b|^2} = \displaystyle \frac{2\pi}{|b|}.

green line

So far, I’ve evaluated the integral Q for the cases |b| = 1 and |b| > 1. Beginning with tomorrow’s post, I’ll evaluate the integral for the case |b| < 1. As it turns out, the method presented above will again be utilized for simplifying the two residues.

1 Comment
by John Quintanilla on August 27, 2015  •  Permalink
Posted in Calculus, Precalculus
Tagged , ,

Posted by John Quintanilla on August 27, 2015

https://meangreenmath.com/2015/08/27/how-i-impressed-my-wife-part-6e/

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

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