How I Impressed My Wife: Part 5a

Amazingly, the integral below has a simple solution:

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|}.

Even more amazingly, the integral Q ultimately does not depend on the parameter a. For several hours, I tried to figure out a way to demonstrate that Q is independent of a, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that Q is independent of a without directly computing the integral Q.

green lineEarlier in this series, I showed that

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

= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}

= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}

= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

I now multiply the top and bottom of this last integral by a^2 + b^2:

Q = \displaystyle \frac{2}{(a^2+b^2)^2} \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }

I now employ the substitution w = (a^2 + b^2) v, so that dw = (a^2 + b^2) v. Since a^2 + b^2 > 0, the endpoints of integration do not change, and so

Q = \displaystyle 2 \int_{-\infty}^{\infty} \frac{dw}{w^2 + b^2 }.

This final integral is independent of a.

green line

Since Q is independent of a, I can substitute any convenient value of a that I wish. For example, I can let a = 0 without altering the value of Q:

Q = \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}

This provides a considerable simplification of the integral Q which also opens up additional methods of evaluation.

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

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