## The Mercator Saga (part 3): the integral

At 60° North, the red box is roughly 10° by 10°. Notice the 2:1 aspect ratio.

You should already have read or skimmed part 1 and part 2 of this saga. In part 2, we showed the picture of the globe (reproduced here) that purported to show how the aspect ratio of a 10° by 10° box at 60° was 2:1.

Of course, it’s only approximately 2:1, I told the students, because (small pause to get attention) the scale on the globe is changing with every degree. In fact, it’s changing continuously. At this point, the students muttered, “mmm, calculus.” So at least they smelt it coming.

I’d like to be able to say that at this point I turned it over to them: “Indeed! Calculus! We’re trying to find $y(\theta)$, the function that gives us the total y-distance on the map as a function of the latitude. Work in your groups to figure out just what calculus you need to do to find that function, and be ready to present in about five minutes.”

The chart, phase three. What happens at an arbitrary angle?

But I can’t. Being realistic, it was good that they sensed calculus, but in this unusual context—the sphere and everything—it would have been excruciating. So I scaffolded like my life depended on it. I reproduce the chart we developed last time that shows the dimensions of the 1° boxes.

“Indeed! Calculus! After all, we want $y(\theta)$, the y-distance as a function of latitude. So to find the total distance, to, say, 60°, I’m going to start at 0° and use this number we have in the table, sec(0°), times 1° because that’s the height per 1°, right?. Then for the next degree, I have to add sec(1°) times 1°, then for the next one, sec(2°) times 1°, all the way up to 60.”

I write the sum on the board, (still ignoring the constant factor, by the way, but that was OK):

$\displaystyle (\sec 0 \times 1) + (\sec 1 \times 1) + (\sec 2 \times 1) + ... + (\sec 59 \times 1)$

“And of course, within each of these 1° sections, it’s changing continuously too. So what should we do?”

Now, almost in unison, “integrate secant,” and a student bravely writes,

$\displaystyle y(\theta) = \int_0^\theta \sec x \,\mathrm{d}x$

### Doing the Integral—the way cool way

“And of course, you all know the integral of the secant, right?”

They squirm, but I take them off the hook. “Of course you don’t. Nobody should remember that. But what’s the most practical way to find this integral?”

“A substitution?” they ask. “Do we, like, do it by parts somehow?”

“All good suggestions, but it’s really not obvious what to substitute. And for parts, usually there’s more stuff, right? So you can have a v and a du? What I’m really asking is, what’s a really practical way to find this integral?”

“Look it up?”

“Yeah! How?”

“That will work, but I want to show you another way.” I pull out my iPhone and hook it up to the projector. “Siri, integrate secant of x.”

“You’re kidding me,” says one kid. Class chatter rises.

“Hmm, let me think…,” says Siri. She shows me an approximation and a link to Wolfram Alpha, which I tap. And this appears:

Siri integrates sec(x)

At this point, they returned to desmos and their data to see if this function actually worked. And it did, to enormous satisfaction throughout the class. That this obscure function, a log of a combination of trig functions, fit these data, was a bit of a miracle, a triumph of actually using calculus.

Data are red dots. Orange is a bogus tangent function. Blue is based on our integral.

### Actually Finding the Integral

We did not find the integral in class, but in case you care, here is a derivation, which shows just how arcane these can get. This is why you look these up. And why it’s so great that Wolfram Alpha exists, because they give you the answer and the derivation.

It turns out that there is a great substitution! First you multiply top and bottom by $(\tan x + \sec x)$:

$\displaystyle \int \sec x \,\mathrm{d}x = \int \frac{\sec x(\tan x + \sec x)}{\tan x + \sec x}\, \mathrm{d}x$

Now we do a u-substitution, choosing

$u = \tan x + \sec x$.

When you take its derivative, you discover that by some miracle,

$\mathrm{d}u = \sec x(\tan x + \sec x)\,\mathrm{d}x$

Which means that

$\displaystyle \int \sec x \,\mathrm{d}x = \int \frac{du}{u} = \ln u = \ln (\tan x + \sec x)$

Which turns out (after even more mind-numbing algebra than it took to find du above) to be equivalent to the horrible formula with the half-angles that Wolfram gave us at first.

Perhaps there is a way to anticipate that that particular substitution would work out so well, but I sure don’t have that kind of insight.