We continue where we left off on the last post about hyperbolic trigonometry. Recall that we ended by finding an antiderivative for using the hyperbolic trigonometric substitution Today, we’ll look at this substitution in more depth.

The functional relationship between and is described by the *gudermannian function,* defined by

This is not at all obvious, so we’ll look at the derivation of this rather surprising-looking formula. It’s the only formula I’m aware of which involves both the arctangent and the exponential function. We remark (as we did in the last post) that we restrict to the interval so that this relationship is in fact invertible.

We use a technique similar to that used to derive a formula for the inverse hyperbolic cosine. First, write

and then multiply through by to obtain the quadratic

This quadratic equation results in

Which sign should we choose? We note that and increase together, so that because is an increasing function of then must be an increasing function of It is not difficult to see that we must choose “plus,” so that and consequently

We remark that no absolute values are required here; this point was discussed in the previous post.

Now to solve for The trick is to use a lesser-known trigonometric identity:

There is such a nice geometrical proof of this identity, I can’t help but include it. Start with the usual right triangle, and extend the segment of length by in order to form an isosceles triangle. Thus,

To find observe that is supplementary to both and so that

which easily implies

Therefore

which is precisely what we need to prove the identity.

Now we substitute back into the previous expression for which results in

This may be solved for giving

So let’s see how to use this to relate circular and hyperbolic trigonometric functions. We have

which after using the usual circular trigonometric identities, becomes

It is also an easy exercise to see that

So revisiting the integral

we may alternatively make the substitution giving

which is the same simple integral we saw in the previous post.

What about the other trigonometric functions? Certainly we know that Again using the usual circular trigonometric identities, we can show that

Knowing these three relationships, the rest are easy to find: and

I think that the gudermannian function should be more widely known. On the face of it, circular and hyperbolic trigonometric functions are *very* different beasts — but they relate to each other in *very* interesting ways, in my opinion.

I will admit that I don’t teach students about the gudermannian function as part of a typical calculus course. Again, there is the issue of time: as you are well aware, students finishing one course in the calculus sequence must be adequately prepared for the next course in the sequence.

So what I do is this: I put the exercises on the gudermannian function as extra challenge problems. Then, if a student is already familiar with hyperbolic trigonometry, they can push a little further to learn about the gudermannian.

Not many students take on the challenge — but there are always one or two who will visit my office hours with questions. Such a treat for a mathematics professor! But I feel it is always necessary to give something to the very best students to chew on, so they’re not bored. The gudermannian does the trick as far as hyperbolic trigonometry is concerned….

As a parting note, I’d like to leave you with a few more exercises which I include in my “challenge” question on the gudermannian. I hope you enjoy working them out!

- Show that
- Show that
- Show that if is the inverse of the gudermannian function, then