How I Impressed My Wife: Part 5j

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} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2}

\displaystyle \int_{-\infty}^{\infty} \left[ \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2} \right] du

if |b| < 1. (The cases |b| = 1 and |b| > 1 have already been handled earlier in this series.)

To complete the calculation, I employ the now-familiar antiderivative

\displaystyle \int \frac{dx}{x^2 + b^2} = \displaystyle \frac{1}{|b|} \tan^{-1} \left( \frac{x}{b} \right).

Using this antiderivative and a simple substitution, I see that

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

= \displaystyle \frac{1}{|b|} \left[ \left( \frac{\pi}{2} + \frac{\pi}{2} \right) - \left( \frac{-\pi}{2} + \frac{-\pi}{2} \right) \right]

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

This completes the fourth method of evaluating the integral Q, using partial fractions.

green line

There’s at least one more way that the integral Q can be calculated, which I’ll begin with tomorrow’s post.

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1 Comment

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

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