How I Impressed My Wife: Part 5g

Earlier in this series, I gave three different methods of showing that

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 \frac{2\pi}{|b|}.

Using the fact that Q is independent of a, I’ll now give a fourth method.
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 \int_{-\infty}^{\infty} \left[ \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_1^2} + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_2^2} \right] du,

where k_1 and k_2 are the positive numbers so that

k_1^2 = 2b^2 + 2|b| \sqrt{b^2-1} - 1,

k_2^2 = 2b^2 - 2|b| \sqrt{b^2-1} - 1,

so that

u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2) (u^2 + k_2^2)

At this point in the calculation, we can employ the now familiar antiderivative

\displaystyle \int \frac{du}{u^2 +k^2} = \displaystyle \frac{1}{k} \tan^{-1} \left( \frac{u}{k} \right) + C,

so that

Q = \left[ \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{k_1}\tan^{-1} \left( \frac{u}{k_1} \right) + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \displaystyle \frac{1}{k_2} \tan^{-1} \left( \frac{u}{k_2} \right) \right]^{\infty}_{-\infty}

= \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{k_1} \left( \frac{\pi}{2} - \frac{-\pi}{2} \right) + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \displaystyle \frac{1}{k_2} \tan^{-1} \left(\frac{\pi}{2} - \frac{-\pi}{2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{2 - 2k_1^2}{k_1} + \frac{2 .k_2^2 - 2}{k_2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{(2 - 2k_1^2)k_2 + (2 k_2^2 - 2)k_1 }{k_1 k_2} \right)

= \displaystyle\frac{\pi}{k_2^2-k_1^2} \left( \frac{2 k_2 - 2k_1^2 k_2 + 2 k_1 k_2^2 - 2 k_1 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{k_2^2-k_1^2} \left( \frac{k_2 - k_1^2 k_2 + k_1 k_2^2 - k_1 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{k_2^2-k_1^2} \left( \frac{k_2 - k_1 - k_1^2 k_2 + k_1 k_2^2 }{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{(k_2-k_1)(k_2+k_1)} \left( \frac{(k_2 - k_1) + k_1 k_2 (k_2 - k_1)}{k_1 k_2} \right)

= \displaystyle\frac{2\pi}{(k_2-k_1)(k_2+k_1)} \left( \frac{(k_2 - k_1)(1+ k_1 k_2)}{k_1 k_2} \right)

= \displaystyle 2\pi \left( \frac{1+ k_1 k_2}{(k_2+k_1) k_1 k_2} \right).

Now it remains to simplify this fraction. To do this, we note that

u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2) (u^2 + k_2^2),

so that

u^4 + (4b^2 - 2)u^2 + 1 = u^4 + (k_1^2 + k_2^2)u^2 + k_1^2 k_2^2

Matching coefficients, we see that

k_1^2 + k_2^2 = 4b^2 -2,

k_1^2 k_2^2 = 1

Since both k_1 and k_2 are positive, we see that k_1 k_2 = 1 from the last equation. Therefore,

k_1^2 + 2 k_1 k_2 + k_2^2 = 4b^2 -2 + 2

(k_1+k_2)^2 = 4b^2

k_1 + k_2 = 2|b|

Plugging these in, we finally conclude that

Q = \displaystyle 2\pi \left( \frac{1+ 1}{2 |b| \cdot 1} \right) = \displaystyle \frac{2\pi}{|b|},

again matching our earlier result.

green line

Using the fourth method, I’ve shown that Q = \displaystyle \frac{2\pi}{|b|} for the cases |b| = 1 and |b| > 1. With tomorrow’s post, I’ll consider the remaining case of |b| < 1.

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1 Comment

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

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