# How I Impressed My Wife: Part 3a

Previously in this series, I showed that $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).
But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.
For this next technique, I begin by using the trigonometric identities $\sin^2 x = \displaystyle \frac{1-\cos 2x}{2}$, $\cos^2x = \displaystyle \frac{1+\cos 2x}{2}$, $2 \sin x \cos x = \sin 2x$.

Using these identities, we obtain $Q = \displaystyle \int_0^{2\pi} \frac{dx}\frac{1+\cos 2x}{2} + a \sin 2x + (a^2 + b^2) \displaystyle \frac{1-\cos 2x}{2}$ $= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

In this way, the exponents have been removed from the denominator, thus making the integrand somewhat less complicated. I’ll continue this different method of evaluating this integral in tomorrow’s post.

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