# How I Impressed My Wife: Part 5f

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

The four roots of the denominator satisfy $u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

In yesterday’s post, I handled the case $|b| = 1$. In today’s post, I’ll consider the case $|b| > 1$, so that $u^2$ is a real number for the four roots of the denominator.

For the sake of simplicity, let me define the positive numbers $k_1$ and $k_2$ so that $k_1^2 = 2b^2 + 2|b| \sqrt{b^2-1} - 1$, $k_2^2 = 2b^2 - 2|b| \sqrt{b^2-1} - 1$.

Clearly $2b^2 + 2|b| \sqrt{b^2-1} - 1 > 0$ if $|b| > 1$, and so we can choose $k_1$ to be positive. For $k_2$, notice that $(2b^2 - 1)^2 = 4b^4 - 4b^2 + 1$,

while $\left[ 2|b| \sqrt{b^2-1} \right]^2 = 4b^2 (b^2 - 1) = 4b^4 - 4b^2$.

Therefore, $(2b^2 - 1)^2 > \left[ 2|b| \sqrt{b^2-1} \right]^2$ $2b^2 - 1 > 2|b| \sqrt{b^2-1}$ $2b^2 - 2|b| \sqrt{b^2-1} - 1 > 0$

So $k_2$ can also be chosen to be a positive number.

Using $k_1$ and $k_2$, I can write $u^4 + (4 b^2 - 2) u^2 + 1 = (u^2 + k_1^2)(u^2 + k_2^2)$,

and so the integrand must have the partial fractions decomposition $\displaystyle \frac{ 2(1+u^2)}{u^4 + (4 b^2 - 2) u^2 + 1} = \displaystyle \frac{A}{u^2 + k_1^2} + \displaystyle \frac{B}{u^2 + k_2^2}$,

Notice that ordinarily, when the denominator contains an irreducible quadratic, the numerator of the partial fractions decomposition has the form $Au + B$ and not $A$. However, there are no $u^3$ and $u$ terms in the denominator, I can treat $u^2$ as the variable for the purposes of the decomposition. Since the right-hand side has linear terms in $u^2$, it suffices to use a constant for finding the decomposition.

To solve for the constants $A$ and $B$, I clear out the denominator: $2u^2 + 2 = A \left[ u^2 + k_2^2 \right] + B \left[ u^2 + k_1^2 \right]$

Matching coefficients, this yields the system of equations $A + B = 2$ $A k_2^2 + B k_1^2 = 2$

Substituting $B = 2-A$ into the second equation, I get $A k_2^2 + (2-A) k_1^2 = 2$ $2 k_1^2 + A (k_2^2 - k_1^2) = 2$ $A (k_2^2 - k_1^2) = 2 - 2k_1^2$ $A = \displaystyle \frac{2 - 2k_1^2}{k_2^2 - k_1^2}$

Therefore, $B = 2 - A = 2 - \displaystyle \frac{2 - 2k_1^2}{k_2^2 - k_1^2}$ $B = \displaystyle \frac{2(k_2^2 - k_1^2) - (2 - 2k_1^2)}{k_2^2 - k_1^2}$ $B = \displaystyle \frac{2 k_2^2 - 2}{k_2^2 - k_1^2}$

Therefore, the integrand has the partial fractions decomposition $\displaystyle \frac{ 2(1+u^2)}{u^4 + (4 b^2 - 2) u^2 + 1} = \displaystyle \left( \frac{2 - 2k_1^2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_1^2} + \displaystyle \left( \frac{2 k_2^2 - 2}{k_2^2 - k_1^2} \right) \frac{1}{u^2 + k_2^2}$ I’ll continue with this fourth evaluation of the integral, continuing the case $|b| > 1$, in tomorrow’s post.