# How I Impressed My Wife: Part 6a

This series was inspired by a question that my wife asked me: calculate $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Originally, I multiplied the top and bottom of the integrand by $\tan^2 x$ and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals. Since $Q$ 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: $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ $= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$ $= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$ $= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$ $= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

Earlier, I evaluated this last integral using partial fractions, separating into the cases $|b| = 1$, $|b| > 1$, and $|b| < 1$. Now, I’ll calculate this same integral using contour integration. (See Wikipedia and Mathworld for more details.) It turns out that $Q$ can be rewritten as $Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$,

where $C_R$ is the contour in the complex plane shown above (graphic courtesy of Mathworld). That’s because $\displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$ $= \displaystyle \lim_{R \to \infty} \int_{-R}^R \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} + \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$ $= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} + \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) du}{z^4 + (4 b^2 - 2) z^2 + 1}$ $= Q + \displaystyle \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$

To show that the limit of the last integral is equal to 0, I use the parameterization $z = R e^{i \theta}$, so that $dz = i R e^{i \theta}$: $\displaystyle \lim_{R \to \infty} \left| \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} \right|$ $= \displaystyle \lim_{R \to \infty} \left| \int_0^{\pi} \frac{ 2R(1+R^2 e^{2i\theta}) d\theta}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|$ $\le \displaystyle \lim_{R \to \infty} \pi \max_{0 \le \theta \le \pi} \left| \frac{ 2R(1+R^2 e^{2i\theta})}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|$ $= \displaystyle \pi \max_{0 \le \theta \le \pi} \lim_{R \to \infty} \left| \frac{ 2R(1+R^2 e^{2i\theta})}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|$ $= \displaystyle \pi \max_{0 \le \theta \le \pi} 0$ $= 0$.

The above limit is equal to zero because the numerator grows like $R^3$ while the denominator grows like $R^4$. (This can be more laboriously established using L’Hopital’s rule).

Therefore, I have shown that $Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}$,

and this contour integral can be computed using residues. I’ll continue with this fifth evaluation of the integral, starting with the case $|b| = 1$, in tomorrow’s post.