How I Impressed My Wife: Part 2b

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

In this series, I’ll explore different ways of evaluating this lineIn yesterday’s post, 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}

We now multiply the top and bottom of the integrand by \sec^2 x. This is permissible because \sec^2 x is defined on the interior of the interval (-\pi/2, \pi/2) — which is why I needed to adjust the limits of integration in the first place. I obtain

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{\cos^2 x \sec^2 x + 2 a \sin x \cos x \sec^2 x + (a^2 + b^2) \sin^2 x \sec^2 x}
Next, I use some trigonometric identities to simplify the denominator:
  • \cos^2 x \sec^2 x = \cos^2 x \displaystyle \frac{1}{\cos^2 x} = 1
  • \sin x \cos x \sec^2 x = \sin x \cos x \frac{1}{\cos^2 x} = \displaystyle \frac{\sin x}{\cos x} = \tan x
  • \sin^2 x \sec^2 x = \sin^2 x \displaystyle \frac{1}{\cos^2 x} = \displaystyle \left( \frac{\sin x}{\cos x} \right)^2 = \tan^2 x

Therefore, the integral becomes

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

green line

I’ll continue with the evaluation of this integral in tomorrow’s post.

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