## Hyperbolic Functions

1. Motivation. There are several different motivations, let me give them here.

Motivation One. We had a problem: find a parametric curve which describes the circle. This problem gave us the trigonometric functions as its solution.

(1)$\displaystyle x^{2}-y^{2}=1$.
What parametric curve produces this hyperbola?

Motivation Two. Recall we defined the trigonometric functions using $\exp(\mathrm{i}t)$ and $\exp(-\mathrm{i}t)$ (and their linear combinations). What happens if we switched these with $\exp(t)$ and $\exp(-t)$ respectively?

2. Parametrizing the Circle. Lets carefully consider how we parametrize the circle. If we draw a unit circle (i.e., a circle with radius 1) in the plane, then consider a slice of it:

What’s the area of the shaded region? Well, the total area is $\pi r^{2}$, and the portion is $\theta/2\pi$. So the slice’s area would be
(2)$\displaystyle A=\left(\frac{\theta}{2\pi}\right)\pi\cdot1^{2}=\frac{\theta}{2}$.
But look, the point on the circle which is drawn lies at $(x,y)=(\cos(\theta),\sin(\theta))$. So in other words,
(2′)$\displaystyle \theta=2A$.
This is how we specify the parameter. If the point is above the $x$ axis, then the area is positive (as in our doodle); if we picked a point on the circle below the $x$ axis, the area would be considered “negative”.

3. Hyperbola. So we consider the “unit” hyperbola $x^{2}-y^{2}=1$ which has its graph be

Then we consider the same song-and-dance: draw a straight line from the origin, find the point where it intersects our hyperbola, and consider the area between the line and the $x$ axis (and between the $y$ axis and the hyperbola) as doodled thus

The area (the shaded grey region) is $A$. Following our parametrization of the circle, we say that $\alpha=2A$ is the “hyperbolic angle”.

If the intersection point is below the $x$ axis, we say the area is negative and simply change the sign.

We say that the hyperbola is the parametrized curve
(3)$\displaystyle x(t)=\cosh(t)$, and $\displaystyle y(t)=\sinh(t)$
where “cosh” (pronounced kosh) is the hyperbolic cosine funcion, and “sinh” (pronounced sin-sh or cinch) is the hyperbolic sine function. Compare this to the usual trigonometric functions.

4. Lets describe this area using integrals. First we will consider the triangular region formed by the $x$ axis and the intersection point:

As an integral, we can express this area as
(4)$\displaystyle A_{tri}=\int^{a}_{0}mx\,\mathrm{d}x$
where $m=b/a$ is the slope of the hypoteneuse.

But this is too much, we need to subtract out the excess area which we doodle as the shaded region in this diagram:

How do we describe this using integration? The error area is
(5)$\displaystyle A_{err}=\int^{a}_{1}\sqrt{x^{2}-1}\,\mathrm{d}x$
and thus the area we want to find is
(6)$\displaystyle A=A_{tri}-A_{err}=\int^{a}_{0}mx\,\mathrm{d}x-\int^{a}_{1}\sqrt{x^{2}-1}\,\mathrm{d}x$
Now evaluating the integral in Equation (5) is tricky!

5. We calculate Equation (5) and the trick requires a change of variables
(7)$\displaystyle x = \sec(\theta)$ and $\displaystyle \mathrm{d}x = \sec(\theta)\tan(\theta)\,\mathrm{d}\theta$
which gives us
(8)$\displaystyle \int\sqrt{x^{2}-1}\,\mathrm{d}x=\int\sqrt{\sec^{2}(\theta)-1}\sec(\theta)\tan(\theta)\,\mathrm{d}\theta$.
Recall the identity
(9)$\displaystyle \sec^{2}(\theta)-1=\tan^{2}(\theta)$.
We plug this into (8) which gives us [changing the $\sqrt{\sec^{2}\theta-1}$ to $\tan(\theta)$]
(10a)$\displaystyle \int\sqrt{x^{2}-1}\,\mathrm{d}x=\int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta$
and using (9) to change $\tan^{2}\theta$ into $\sec^{2}\theta-1$ produces
(10b)$\displaystyle \int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\int\sec^{3}(\theta)-\sec(\theta)\,\mathrm{d}\theta$.
What now?

We consider the integral
$\displaystyle \int\sec^{3}(\theta)\,\mathrm{d}\theta$.
What to do? We integrate by parts, writing
(11a)$f'(\theta)=\sec^{2}(\theta)$ thus $\displaystyle f(\theta)=\tan(\theta)$
and
(11b)$g(\theta)=\sec(\theta)$ thus $\displaystyle g'(\theta)=\sec(\theta)\tan(\theta)$
We write our integral as
(11c)$\displaystyle \int\sec^{3}(\theta)\,\mathrm{d}\theta = \int g(\theta)f'(\theta)\,\mathrm{d}\theta$.
Integrating by parts gives us
(12)$\displaystyle \int\sec^{3}(\theta)\,\mathrm{d}\theta = \tan(\theta)\sec(\theta)-\int \sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta$.
Does this seem familiar?

We plug (12) into the right hand side of (10b) producing
(13a)$\displaystyle \int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\tan(\theta)\sec(\theta)-\int \sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta-\int\sec(\theta)\,\mathrm{d}\theta$.
We add $\int\tan^{2}\theta\sec\theta\,\mathrm{d}\theta$ to both sides
(13b)$\displaystyle 2\int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\tan(\theta)\sec(\theta)-\int\sec(\theta)\,\mathrm{d}\theta$.
Divide both sides by 2
(13c)$\displaystyle \int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\frac{\tan(\theta)\sec(\theta)}{2}-\frac{1}{2}\int\sec(\theta)\,\mathrm{d}\theta$.
Now we just need to evaluate $\int\sec\theta\,\mathrm{d}\theta$!

Integrating secant is a little tricky. First we write
(14a)$\displaystyle \int\sec(\theta)\,\mathrm{d}\theta=\int\sec(\theta)\frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}\,\mathrm{d}\theta$.
Then we consider a change of variables
(14b)$\displaystyle u = \sec(\theta)+\tan(\theta)$
then we find
$\displaystyle\mathrm{d}u = (\sec(\theta)\tan(\theta) + \sec^{2}(\theta))\,\mathrm{d}\theta$.

Thus we have
(14b)$\displaystyle \int\sec(\theta)\frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}\,\mathrm{d}\theta = \int\frac{\mathrm{d}u}{u}$.
We can evaluate this quickly as
(14c)$\displaystyle \int\frac{\mathrm{d}u}{u} = \ln(u)+C$.
Now we unroll our substitution, and write this as
(14d)$\displaystyle \int\frac{\mathrm{d}u}{u} = \ln(\sec(\theta)+\tan(\theta))+C$.
Thus our integral (14a) is
(14a’)$\displaystyle \int\sec(\theta)\,\mathrm{d}\theta=\ln(\sec(\theta)+\tan(\theta))+C$.
Great, so we can recall (13c) is
(13c)$\displaystyle \int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\frac{\tan(\theta)\sec(\theta)}{2}-\frac{1}{2}\int\sec(\theta)\,\mathrm{d}\theta$.
Plugging in (14a’) to the right hand side gives us
(15)$\displaystyle \int\sec(\theta)\tan^{2}(\theta)\,\mathrm{d}\theta=\frac{\tan(\theta)\sec(\theta)}{2}-\frac{1}{2}\ln(\sec(\theta)+\tan(\theta))+C$.
What now?

Well, we have to keep unrolling our substitution. We had used $x=\sec(\theta)$, so
$\displaystyle\tan(\theta)=\sqrt{\sec^{2}(\theta)-1}=\sqrt{x^{2}-1}$.
Now what? Well, we use these and rewrite Equation (15) as
(16)$\displaystyle \frac{\sec(\theta)\tan(\theta)}{2}-\frac{1}{2}\ln(\sec(\theta)+\tan(\theta))+C = \frac{x\sqrt{x^{2}-1}}{2}-\frac{1}{2}\ln(x+\sqrt{x^{2}-1})+C$.
We keep on unrolling, and find the solution to the indefinite integral
(16)$\displaystyle \int\sqrt{x^{2}-1}\,\mathrm{d}x = \frac{x\sqrt{x^{2}-1}}{2}-\frac{1}{2}\ln(x+\sqrt{x^{2}-1})+C$.
My goodness, that was a lot of algebra!

Are we done yet? No! We still have to compute the error area, which is
(17)$\displaystyle \int^{a}_{1}\sqrt{x^{2}-1}\,\mathrm{d}x = \left.\frac{x\sqrt{x^{2}-1}}{2}-\frac{1}{2}\ln(x+\sqrt{x^{2}-1})\right|^{a}_{1}$.
After some fumbling arithmetic, we find
(18)$\displaystyle \int^{a}_{1}\sqrt{x^{2}-1}\,\mathrm{d}x = \frac{a\sqrt{a^{2}-1}}{2}-\frac{1}{2}\ln(a+\sqrt{a^{2}-1})$.
That gives us the error area.

6. So now we have the error area, and the triangle’s area recall is
(4)$\displaystyle A_{tri}=\int^{a}_{0}mx\,\mathrm{d}x$
where $m=b/a$ is the slope of the hypoteneuse. So we have
(19)$\displaystyle A_{tri}=\frac{ma^{2}}{2}=\frac{ab}{2}$.
So the total area is
(20)$\displaystyle A=A_{tri}-A_{err}=\frac{ab}{2}-\frac{a\sqrt{a^{2}-1}}{2}+\frac{1}{2}\ln(a+\sqrt{a^{2}-1})$.
Recall that the point $(a,b)=(a,\sqrt{a^{2}-1})$, so we simplify this as
(20′)$\displaystyle A=\frac{1}{2}\ln(a+\sqrt{a^{2}-1})$.
That’s the area!

Why were we looking for this? Well, the hyperbolic angle (which we were describing in terms of the area) is
(21)$\displaystyle \alpha=2A$.
Observe, using Equation (20′), we have the hyperbolic angle be
(22)$\displaystyle \alpha=\ln(a+\sqrt{a^{2}-1})$.
How, oh how, can we find the parametric form of the hyperbola with this puny angle?

The first step is to rewrite (22) with $a=\sqrt{b^{2}+1}$, giving us
(22′)$\displaystyle \alpha=\ln(\sqrt{b^{2}+1}+b)$.
We introduce hyperbolic “trigonometric” functions to parametrize the hyperbola:
(23a)$\displaystyle \cosh\left(\ln(\sqrt{b^{2}+1}+b)\right)=a=\sqrt{1+b^{2}}$
and
(23b)$\displaystyle \sinh\left(\ln(\sqrt{b^{2}+1}+b)\right)=b$.
What are the properties of these functions? And what are they, anyways?

Exercise 1: prove the identity $\cosh^{2}(x)-\sinh^{2}(x)=1$.

From this exercise, we may “guess” that the parametric curve describing a hyperbola is
(24)$\displaystyle x(t)=\cosh(t)$ and $\displaystyle y(t)=\sinh(t)$.
We observe that, if
$\displaystyle \alpha=\ln(\sqrt{b^{2}+1}+b)$
then
(25a)$\displaystyle \exp(\alpha)=\sqrt{b^{2}+1}+b$
and, after some algebraic manipulations,
(25b)$\displaystyle \exp(-\alpha)=\sqrt{b^{2}+1}-b$.
Thus we find
(26a)$\displaystyle \exp(\alpha)+\exp(-\alpha)=2\sqrt{b^{2}+1}=2a$
and
(26b)$\displaystyle \exp(\alpha)-\exp(-\alpha)=2b$.
Thus we deduce
(27a)$\displaystyle \cosh(\alpha)=\frac{\exp(\alpha)+\exp(-\alpha)}{2}$
and
(27b)$\displaystyle \sinh(\alpha)=\frac{\exp(\alpha)-\exp(-\alpha)}{2}$.
Just as we prophesized earlier on!

Exercise 2. Find the derivatives of $\sinh(x)$ and $\cosh(x)$.

Exercise 3. Construct the hyperbolic tangent “in the obvious way” and find its derivative.

Exercise 4. Construct the hyperbolic secant and hyperbolic cosecant, find their derivatives.

Exercise 5. Compute $\int\sinh(x)\,\mathrm{d}x$ and $\int\cosh(x)\,\mathrm{d}x$.

Exercise 6. Calculate $\int\tanh(x)\,\mathrm{d}x$.

References. I acknowledge using Isaac Greenspan’s Deriving the Hyperbolic Trigonometric Functions [pdf].