# My Favorite One-Liners: Part 69

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This story, that I’ll share with my Precalculus students, comes from Fall 1996, my first semester as a college professor. I was teaching a Precalculus class, and the topic was vectors. I forget the exact problem (believe me, I wish I could remember it), but I was going over the solution of a problem that required finding $\tan^{-1}(7)$. I told the class that I had worked this out ahead of time, and that the approximate answer was $82^o$. Then I used that angle for whatever I needed it for and continued until obtaining the eventual solution.

(By the way, I now realize that I was hardly following best practices by computing that angle ahead of time. Knowing what I know now, I should have brought a calculator to class and computed it on the spot. But, as a young professor, I was primarily concerned with getting the answer right, and I was petrified of making a mistake that my students could repeat.)

After solving the problem, I paused to ask for questions. One student asked a good question, and then another.

Then a third student asked, “How did you know that $\tan^{-1}(7)$ was $82^o$?

The class immediately erupted… some with laughter, some with disbelief. (I had a terrific rapport with those students that semester; part of the daily atmosphere was the give-and-take with any number of exuberant students.) One guy in the front row immediately challenged me: “Oh yeah? Then what’s $\tan^{-1}(9)$?

I started to stammer, “Uh, um…”

“Aha!” they said. “He’s faking it.” They start pulling out their calculators.

Then I thought as fast as I could. Then I realized that I knew that $\tan 82^o \approx 7$, thanks to my calculation prior to class. I also knew that $\displaystyle \lim_{x \to 90^-} \tan x = \infty$ since the graph of $y = \tan x$ has a vertical asymptote at $x = \pi/2 = 90^o$. So the solution to $\tan x = 9$ had to be somewhere between $82^o$ and $90^o$.

So I took a total guess. “$84^o$,” I said, faking complete and utter confidence.

Wouldn’t you know it, I was right. (The answer is about $83.66^o$.)

In stunned disbelief, the guy who asked the question asked, “How did you do that?”

I was reeling in shock that I guessed correctly. But I put on my best poker face and answered, “I told you, I had it memorized.” And then I continued with the next example. For the rest of the semester, my students really thought I had it memorized.

To this day, this is my favorite stunt that I ever pulled off in front of my students.

# Engaging students: Inverse trigonometric functions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Joe Wood. His topic, from Precalculus: inverse trigonometric functions.

What are the contributions of various cultures to this topic?

Trig functions have a very long history spanning many countries and cultures. Greek astronomers such as Aristarchus, Claudius, and Ptolemy first used trigonometry; however, according to the University of Connecticut, these Greek astronomers were primarily concerned with “the length of the chord of a circle as a function of the circular arc joining its endpoints.” Many of these astronomers, Ptolemy especially, were concerned with planetary and celestial body’s rotations, so this made sense.

While the Greeks first studied trigonometric concepts, it was the Indian people who really studied sine and cosine functions with the angle as a variable. The information was then brought to the Arabic and Persian cultures. One significant figure, a Persian by the name Abu Rayhan Biruni, used trig to accurately estimate the circumference of Earth and its radius before the end of the 11th century.

Fast-forward about 700 years, a Swiss mathematician, Daniel Bernoulli, used the “A.sin” notation to represent the inverse of sine. Shortly after, another Swiss mathematician used “A t” to represent the inverse of tangent. That man was none other than Leonhard Euler.  It was not until 1813 that the notation sin-1 and tan-1 were introduced by Sir John Fredrick William Herschel, an English mathematician.

As we can see, the development of inverse trigonometric functions took quite the cultural rollercoaster ride before stopping some place we see being familiar. It took many cultures, and even more years to develop this sophisticated branch of mathematics.

How could you as a teacher create an activity or project that involves your topic?

Last Semester I taught a lesson on the trigonometric identities. I found this cool cut and paste activity for the students that allowed them to warm up to the trig identities by not having to do the process themselves, but still having to see every step of converting one trig function into another with the identities. Below, you will find the activity, then the instructions, and finally how to modify the activity to fit inverse trig identities specifically.

Directions:
1.) Begin by cutting out all the pieces.
2.) Students will take any of the four puzzle pieces with the black squiggly line.
3.) Find an equivalent puzzle piece by using some trig identity.
4.) Repeat step 3 until there are no more equivalent pieces.
5.) Grab the next puzzle pieces with the black squiggly line.
6.) Repeat steps 3-5 until all puzzle pieces have been used.
Ex.) Begin with cscx-sinx. Lay next to that piece, the piece that reads =1/sinx – sinx, then the piece that reads =1/sinx – sin2x/sinx. Contiue the trend until you reach =cotx * cosx. Then move to the next squiggly lined piece.

Modify:
This game can be modified using inverse trig functions. Start with pieces such as sin-1(sin(300)) in squiggles. Have a piece showing sin-1(sin(300)) with a line through the sines. Then a piece that just shows 300. Next a piece in a squiggly line that is sin-1(sqrt(2)/2) that connects to a piece of 450, but make them write why this works.

How does this topic extend what your students should have learned in previous courses?

Obviously, by this time students should know what trigonometric functions are and how to use this. Students should also know from previous classes what inverse functions are. Studying inverse trig functions then is a continuation of these topics. As I teacher I would begin relating inverse trig functions by refreshing the students on what inverse functions are. The class would then move into the concept that the trig expression of an angle returns a ratio of two sides of a triangle. We would slowly move into what happens then if you know the sides of a triangle but need the angle. From there we would discuss trigonometric expressions using the angles as variables. Finally, we would make the connection that that is a function, and on the proper interval should have an inverse function. That is when the extension into the new topic of inverse trigonometric functions would seriously begin.

# 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.

# How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

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

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction
Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by $\sec^2 x$, and the substitution $u = \tan x$.
Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution $u = \tan \theta/2$.
Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle
Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter $a$, the magic substitution $u = \tan \theta/2$, and partial fractions.
Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter $a$, the magic substitution $u = \tan \theta/2$, and contour integration using the real line and an expanding semicircle.
Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

# The antiderivative of 1/(x^4+1): Part 10

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

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

Also, as long as $x \ne 1$ and $x \ne -1$, there is an alternative answer:

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

In this concluding post of this series, I’d like to talk about the practical implications of the assumptions that $x \ne 1$ and $x \ne -1$.

For the sake of simplicity for the rest of this post, let

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

and

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

If I evaluate a definite integral of $\displaystyle \frac{1}{x^4+1}$ over an interval that contains neither $x = 1$ or $x = -1$, then either $F$ or $G$ can be used. Courtesy of Mathematica:

However, if the region of integration contains either $x = -1$ or $x =1$ (or both), then only using $F$ returns the correct answer.

So this should be a cautionary tale about solving for angles, as the innocent-looking $+n\pi$ that appeared several posts ago ultimately makes a big difference in the final answers that are obtained.

# 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

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

or

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

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.

# The antiderivative of 1/(x^4+1): Part 6

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

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

So far, I’ve shown that the denominator can be factored over the real numbers:

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

$= \displaystyle \int \frac{ -\displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx + \int \frac{ \displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx$

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

To evaluate the remaining two integrals, I’ll use the antiderivative

$\displaystyle \int \frac{dx}{x^2 + k^2} = \displaystyle \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right)$.

To begin, I’ll complete the squares:

$\displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx = \displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} }$

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

Applying the substitutions $u = x - \displaystyle \frac{ \sqrt{2}}{2}$ and $v = x + \displaystyle \frac{ \sqrt{2}}{2}$, I can continue:

$= \displaystyle \frac{1}{4} \int \frac{ du }{ u^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 } + \frac{1}{4} \int \frac{ dv }{v^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 }$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{u}{\sqrt{2}/2} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{v }{\sqrt{2}/2} \right) + C$

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

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

Combining, I finally arrive at the answer for $\displaystyle \int \frac{dx}{x^4 + 1}$:

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

Naturally, this can be checked by differentiation, but I’m not going type that out.

# Inverse Functions: Arctangent and Angle Between Two Lines (Part 25)

The smallest angle between the non-perpendicular lines $y = m_1 x + b_1$ and $y = m_1 x + b_2$ can be found using the formula

$\theta = \displaystyle \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$.

A generation ago, this formula used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry). However, I find that analytic geometry has fallen out of favor in modern Precalculus courses.

Why does this formula work? Consider the graphs of $y = m_1 x$ and $y = m_1 x + b_1$, and let’s measure the angle that the line makes with the positive $x-$axis.

The lines $y = m_1 x + b_1$ and $y = m_1 x$ are parallel, and the $x-$axis is a transversal intersecting these two parallel lines. Therefore, the angles that both lines make with the positive $x-$axis are congruent. In other words, the $+ b_1$ is entirely superfluous to finding the angle $\theta_1$. The important thing that matters is the slope of the line, not where the line intersects the $y-$axis.

The point $(1, m_1)$ lies on the line $y = m_1 x$, which also passes through the origin. By definition of tangent, $\tan \theta_1$ can be found by dividing the $y-$ and $x-$coordinates:

$\tan \theta_1 = \displaystyle \frac{m}{1} = m_1$.

We now turn to the problem of finding the angle between two lines. As noted above, the $y-$intercepts do not matter, and so we only need to find the smallest angle between the lines $y = m_1 x$ and $y = m_2 x$.

The angle $\theta$ will either be equal to $\theta_1 - \theta_2$ or $\theta_2 - \theta_1$, depending on the values of $m_1$ and $m_2$. Let’s now compute both $\tan (\theta_1 - \theta_2)$ and $\tan (\theta_2 - \theta_1)$ using the formula for the difference of two angles:

$\tan (\theta_1 - \theta_2) = \displaystyle \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2}$

$\tan (\theta_2 - \theta_1) = \displaystyle \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_2 \tan \theta_1}$

Since the smallest angle $\theta$ must lie between $0$ and $\pi/2$, the value of $\tan \theta$ must be positive (or undefined if $\theta = \pi/2$… for now, we’ll ignore this special case). Therefore, whichever of the above two lines holds, it must be that

$\tan \theta = \displaystyle \left| \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} \right|$

We now use the fact that $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$:

$\tan \theta = \displaystyle \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

$\theta = \tan^{-1} \left( \displaystyle \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$

The above formula only applies to non-perpendicular lines. However, the perpendicular case may be remembered as almost a special case of the above formula. After all, $\tan \theta$ is undefined at $\theta = \pi/2 = 90^\circ$, and the right hand side is also undefined if $1 + m_1 m_2 = 0$. This matches the theorem that the two lines are perpendicular if and only if $m_1 m_2 = -1$, or that the slopes of the two lines are negative reciprocals.