How I Impressed My Wife: Part 6d

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:


Fortunately, humans can still do some things that computers can’t. As observed yesterday, 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, so that

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

r_1 r_2 = 1.

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{1-r_1^2}{r_1 (-2r_1^2 + 4b^2-2)} + \frac{1-r_2^2}{r_2(-2r_2^2 + 4b^2-2)} \right]

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

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

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

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

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

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

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

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

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

= \displaystyle \frac{4\pi}{r_1 + r_2}

To complete the calculation, I observe that

(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|.


Q = \displaystyle \frac{4\pi}{r_1 + r_2} = \displaystyle \frac{4\pi}{2|b|} = \displaystyle \frac{2\pi}{|b|}.

green line

In tomorrow’s post, I’ll present another way to simplify this nasty algebraic expression.

One thought on “How I Impressed My Wife: Part 6d

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