SOHCAHTOA

Years ago, when I first taught Precalculus at the college level, I was starting a section on trigonometry by reminding my students of the acronym SOHCAHTOA for keeping the trig functions straight:

$\sin \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Hypotenuse}}$ ,

$\cos \theta = \displaystyle \frac{\hbox{Adjacent}}{\hbox{Hypotenuse}}$ ,

$\tan \theta = \displaystyle \frac{\hbox{Opposite}}{\hbox{Adjacent}}$ .

At this point, one of my students volunteered that a previous math teacher had taught her an acrostic to keep these straight: Some Old Hippie Caught Another Hippie Tripping On Acid.

Needless to say, I’ve been passing this pearl of wisdom on to my students ever since.

10 Secret Trig Functions Your Math Teachers Never Taught You

Students in trigonometry are usually taught about six functions:

$\sin \theta, \cos \theta, \tan \theta, \cot \theta, \sec \theta, \csc \theta$

I really enjoyed this article about trigonometric functions that were used in previous generations but are no longer taught today, like $\hbox{versin} \theta$ and $\hbox{havercosin} \theta$ :

http://blogs.scientificamerican.com/roots-of-unity/10-secret-trig-functions-your-math-teachers-never-taught-you/

Naturally, Math With Bad Drawings had a unique take on this by adding a few more suggested functions to the list. My favorites:

Hippopotenuse

Source: https://www.facebook.com/sandraboynton/photos/a.206184542749790.56269.114554295246149/1075341509167418/?type=3&theater

The antiderivative of 1/(x^4+1): Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

$\displaystyle \int \frac{dx}{x^4+1}$

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

Different Ways of Solving a Contest Problem: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of solving the contest problem “If $3 \sin \theta = \cos \theta$ , what is $\sin \theta \cos \theta$ ?”

Part 1: Drawing the angle $\theta$

Part 2: A first attempt using a Pythagorean identity.

Part 3: A second attempt using a Pythagorean identity and the original hypothesis for $\theta$ .

The antiderivative of 1/(x^4+1): Part 9

In the course of evaluating the antiderivative

$\displaystyle \int \frac{1}{x^4 + 1} dx$ ,

I have stumbled across a very curious trigonometric identity:

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < x_2$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $x_2 < x < x_1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> x_1$ ,

where $x_1$ and $x_2$ are the unique values so that

$\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ ,

$\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}$ .

I will now show that $x_1 = 1$ and $x_2 = -1$ . Indeed, it’s apparent that these have to be the two transition points because these are the points where $\displaystyle \frac{x \sqrt{2}}{1 - x^2}$ is undefined. However, it would be more convincing to show this directly.

To show that $x_1 = 1$ , I need to show that

$\tan^{-1} (\sqrt{2} - 1 ) + \tan^{-1}( \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ .

I could do this with a calculator…

…but that would be cheating.

Instead, let $\alpha = \tan^{-1} (\sqrt{2} - 1 )$ and $\beta = \tan^{-1} (\sqrt{2} + 1 )$ , so that

$\tan \alpha = \sqrt{2} - 1$ ,

$\tan \beta = \sqrt{2} + 1$ .

Indeed, by SOHCAHTOA, the angles $\alpha$ and $\beta$ can be represented in the figure below:

The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly $\sqrt{2} - 1 + \sqrt{2} + 1 = 2\sqrt{2}$ . I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing $\alpha$ , the missing side is

$\sqrt{ \left(\sqrt{2} - 1 \right)^2 + 1^2} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4-2\sqrt{2}}$

Next, for the small right triangle containing $\beta$ , the missing side is

$\sqrt{ \left(\sqrt{2} + 1 \right)^2 + 1^2} = \sqrt{2 + 2\sqrt{2} + 1 + 1} = \sqrt{4+2\sqrt{2}}$

So let me redraw the figure, eliminating the altitude from the previous figure:

Notice that the condition of the Pythagorean theorem is satisfied, since

$\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = 4 - 2\sqrt{2} + 4 + 2 \sqrt{2} = 8$ ,

$\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = \left( 2\sqrt{2} \right)^2$ .

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so $\alpha + \beta = \pi/2$ . In other words, $x_1 = 1$ , as required.

To show that $x_2 = -1$ , I will show that the function $f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ is an odd function using the fact that $\tan^{-1} x$ is also an odd function:

$f(-x) = \tan^{-1} ( -x\sqrt{2} - 1 ) + \tan^{-1}( -x \sqrt{2} + 1)$

$= \tan^{-1} ( -[x\sqrt{2} + 1] ) + \tan^{-1}( -[x \sqrt{2} - 1])$

$= -\tan^{-1} ( x\sqrt{2} + 1 ) - \tan^{-1}( x \sqrt{2} - 1)$

$= - \left[ \tan^{-1} ( x\sqrt{2} + 1 ) + \tan^{-1}( x \sqrt{2} - 1) \right]$

$= -f(x)$ .

Therefore, $f(-1) = -f(1) = -\displaystyle \frac{\pi}{2}$ , and so $x_2 = -1$ .

The antiderivative of 1/(x^4+1): Part 8

In the course of evaluating the antiderivative

$\displaystyle \int \frac{1}{x^4 + 1} dx$ ,

I’ve accidentally stumbled on a very curious looking trigonometric identity:

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < -1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $-1 < x < 1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> 1$ .

The extra $-\pi$ and $\pi$ are important. Without them, the graphs of the left-hand side and right-hand sides are clearly different if $x < -1$ or $x > 1$ :

However, they match when those constants are included:

Let’s see if I can explain why this trigonometric identity occurs without resorting to the graphs.

Since $\tan^{-1} x$ assumes values between $-\pi/2$ and $\pi/2$ , I know that

$-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) < \frac{\pi}{2}$ ,

$-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} + 1 ) < \frac{\pi}{2}$ ,

and so

$-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi$ .

However,

$-\displaystyle \frac{\pi}{2} < \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2}$ ,

and so $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )$ and $\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ must differ if $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1)$ is in the interval $[-\pi,-\pi/2]$ or in the interval $[\pi/2,\pi]$ .

I also notice that

$-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi$ ,

$-\displaystyle \frac{\pi}{2} < -\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2}$ ,

and so

$-\displaystyle \frac{3\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )-\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{3\pi}{2}$ .

However, this difference can only be equal to a multiple of $\pi$ , and there are only three multiples of $\pi$ in the interval $\displaystyle \left( -\frac{3\pi}{2}, \frac{3\pi}{2} \right)$ , namely $-\pi$ , $0$ , and $\pi$ .

To determine the values of $x$ where this happens, I also note that $f_1(x) = x \sqrt{2} - 1$ , $f_2(x) = x \sqrt{2} + 1$ , and $f_3(x) = \tan^{-1} x$ are increasing functions, and so $f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ must also be an increasing function. Therefore, to determine where $f(x)$ lies in the interval $[\pi/2,\pi]$ ,it suffices to determine the unique value $x_1$ so that $f(x_1) = \pi/2$ . Likewise, to determine where $f(x)$ lies in the interval $[-\pi,-\pi/2]$ ,it suffices to determine the unique value $x_2$ so that $f(x_2) = -\pi/2$ .

In summary, I have shown so far that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < x_2$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $x_2 < x < x_1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> x_1$ ,

where $x_1$ and $x_2$ are the unique values so that

$\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ ,

$\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}$ .

So, to complete the proof of the trigonometric identity, I need to show that $x_1 = 1$ and $x_2 = -1$ . I will do this in tomorrow’s post.

The antiderivative of 1/(x^4+1): Part 7

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

As we’ve seen in this series, the answer is

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

It turns out that this can be simplified somewhat as long as $x \ne 1$ and $x \ne -1$ . I’ll use the trig identity

$\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

When I apply this trig identity for $\alpha = \tan^{-1} ( x\sqrt{2} - 1 )$ and $\beta = \tan^{-1} ( x\sqrt{2} + 1 )$ , I obtain

$\tan \left[ \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) \right] = \displaystyle \frac{x \sqrt{2} - 1 + x \sqrt{2} + 1}{1 - (x\sqrt{2} - 1)(x\sqrt{2} + 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{1 - (2x^2 - 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{2 - 2x^2}$

$= \displaystyle \frac{x \sqrt{2}}{1 - x^2}$ .

So we can conclude that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + n\pi$

for some integer $n$ that depends on $x$ . The $+n\pi$ is important, as a cursory look reveals that $y = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ and $y = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when $x = 1$ or $x = -1$ .

The two graphs coincide when $-1 < x < 1$ but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract $\pi$ from the orange graph if $x < -1$ and add $\pi$ to the orange graph if $x > 1$ , then they match:

So, evidently

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < -1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $-1 < x < 1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> 1$ .

So as long as $x \ne 1$ and $x \ne -1$ , this constant $-\pi$ , $0$ , or $\pi$ can be absorbed into the constant $C$ :

$\displaystyle \int \frac{1}{x^4 + 1} dx = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + C$ .

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.

Different ways of solving a contest problem (Part 1)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$ , what is $\sin \theta \cos \theta$ ?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Here’s the first solution that I received: draw the appropriate triangles for the angle $\theta$ :

$3 \sin \theta = \cos \theta$

$\tan \theta = \displaystyle \frac{1}{3}$

Therefore, the angle $\theta$ must lie in either the first or third quadrant, as shown. (Of course, $\theta$ could be coterminal with either displayed angle, but that wouldn’t affect the values of $\sin \theta$ or $\cos \theta$ .)

In Quadrant I, $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ and $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$ . Therefore,

$\sin \theta \cos \theta = \displaystyle \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \displaystyle \frac{3}{10}$ .

In Quadrant III, $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ and $\cos \theta = -\displaystyle \frac{3}{\sqrt{10}}$ . Therefore,

$\sin \theta \cos \theta = \displaystyle \left( - \frac{1}{\sqrt{10}} \right) \times \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$ .

Either way, we can be certain that $\sin \theta \cos \theta = \displaystyle \frac{3}{10}$ .

How I Impressed My Wife: Part 5b

Amazingly, the integral below has a simple solution:

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

Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$ . For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$ , but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$ .

Earlier in this series, I showed that

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

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

$= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

Yesterday, I showed used the substitution $w = (a^2 + b^2) v$ to show that $Q$ was independent of $a$ . Today, I’ll use a different method to establish the same result. Let

$Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }$ .

Notice that I’ve written this integral as a function of the parameter $a$ . I will demonstrate that $Q'(a) = 0$ , so that $Q(c)$ is a constant with respect to $a$ . In other words, $Q(a)$ does not depend on $a$ .

To do this, I differentiate under the integral sign with respect to $a$ (as opposed to $x$ ) using the Quotient Rule:

$Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

I now apply the trigonometric substitution $v = \displaystyle \frac{b}{a^2+b^2} \tan \theta$ , so that

$(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta$

and

$dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

The endpoints of integration change from $-\infty < v < \infty$ to $-\pi/2 < \theta < \pi/2$ , and so

$Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta$

$= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]$

$= 0$ .

I’m not completely thrilled with this demonstration that $Q$ is independent of $a$ , mostly because I had to do so much simplification of the integral $Q$ to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

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

and demonstrate that $Q$ is independent of $a$ , perhaps by differentiating $Q$ with respect to $a$ and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying $Q$ first.