# How I Impressed My Wife: Part 5i

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. 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: $Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$ $= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$ $= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$ $= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$ $= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$ $= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2}$

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

I will evaluate this integral using partial fractions. The denominator factors as the product of two irreducible quadratics, and so I must solve $\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{Cu + D}{[u + \sqrt{1-b^2}]^2 +b^2}$,

or $\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{Au + B}{u^2 - 2u \sqrt{1-b^2} +1} + \frac{Cu + D}{u^2 + 2u\sqrt{1-b^2} + 1}$.

I now clear out the denominators: $2(1+u^2) = (Au+B)(u^2 + 2u\sqrt{1-b^2} + 1) + (Cu+D)(u^2 - 2u\sqrt{1-b^2} + 1)$

Now I multiply out the right-hand side: $Au^3 + 2Au^2 \sqrt{1-b^2} + Au + Bu^2 + 2Bu\sqrt{1-b^2} + B$ $+ Cu^3 - 2Cu^2 \sqrt{1-b^2} + Cu + Du^2 - 2Du\sqrt{1-b^2} + D$

Equating this with $2u^2 + 2$ and matching coefficients yields the following system of four equations in four unknowns: $A + C = 0$ $2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$ $Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$ $B + D = 2$

Ordinarily, four-by-four systems of equations are rather painful to solve, but this system isn’t so bad.

From the first equation, I see that $C = -A$.

From the third equation, I see that $Au + 2Bu \sqrt{1-b^2} + Cu - 2Du \sqrt{1-b^2} = 0$ $Au + 2Bu \sqrt{1-b^2} -Au - 2Du \sqrt{1-b^2} = 0$ $2Bu \sqrt{1-b^2} - 2Du \sqrt{1-b^2} = 0$ $B - D = 0$ $B = D$.

From the fourth equation, I see that $B + D = 2$ $B + B = 2$ $2B = 2$ $B = 1$,

so that $D =1$ as well. Finally, from the second equation, I see that $2A\sqrt{1-b^2} + B - 2C \sqrt{1-b^2} + D = 2$ $2A\sqrt{1-b^2} + 1 -2(-A) \sqrt{1-b^2} + 1 = 2$ $4A\sqrt{1-b^2} = 0$ $A= 0$,

so that $C = 0$ as well. This yields the partial fractions decomposition $\displaystyle \frac{ 2(1+u^2)}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)} = \displaystyle \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2}$.

This can be confirmed by directly adding the fractions on the right-hand side: $\displaystyle \frac{1}{[u - \sqrt{1-b^2}]^2 +b^2} + \frac{1}{[u + \sqrt{1-b^2}]^2 +b^2}$ $= \displaystyle \frac{[u + \sqrt{1-b^2}]^2 +b^2 + [u - \sqrt{1-b^2}]^2 +b^2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$ $= \displaystyle \frac{u^2 + 2u\sqrt{1-b^2} + 1 - b^2 + b^2 + u^2 - 2u\sqrt{1-b^2} + 1 - b^2 + b^2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$ $= \displaystyle \frac{2u^2 + 2}{([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$. I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$, in tomorrow’s post.