# College Math Teaching

## October 29, 2014

### Hyperbolic Trig Functions and integration…

In college calculus courses, I’ve always wrestled with “how much to cover in the hyperbolic trig functions” section.

On one hand, the hyperbolic trig functions make some integrals much easer. On the other hand: well, it isn’t as if our classes are populated with the highest caliber student (I don’t teach at MIT); many struggle with the standard trig functions. There is only so much that the average young mind can absorb.

In case your memory is rusty:

$cosh(x) =\frac{e^x + e^{-x}}{2}, sinh(x) = \frac{e^x -e^{-x}}{2}$ and then it is immediate that the standard “half/double angle formulas hold; we do remember that $\frac{d}{dx}cosh(x) = sinh(x), \frac{d}{dx} = cosh(x)$.

What is less immediate is the following: $sinh^{-1}(x) = ln(x+\sqrt{x^2+1}), cosh^{-1}(x) = ln(x + \sqrt{x^2 -1}) (x \ge 1)$.

Exercise: prove these formulas. Hint: if $sinh(y) = x$ then $e^{y} - 2x- e^{-y} =0$ so multiply both sides by $e^{y}$ to obtain $e^{2y} -2x e^y - 1 =0$ now use the quadratic formula to solve for $e^y$ and keep in mind that $e^y$ is positive.

For the other formula: same procedure, and remember that we are using the $x \ge 0$ branch of $cosh(x)$ and that $cosh(x) \ge 1$

The following follows easily: $\frac{d}{dx} sinh^{-1} (x) = \frac{1}{\sqrt{x^2 + 1}}$ (just set up $sinh(y) = x$ and use implicit differentiation followed by noting $cosh^2(x) -sinh^2(x) = 1$. ) and $\frac{d}{dx} cosh^{-1}(x) = \frac{1}{\sqrt{x^2-1}}$ (similar derivation).

Now, we are off and running.

Example: $\int \sqrt{x^2 + 1} dx =$

We can make the substitution $x =sinh(u), dx = cosh(u) du$ and obtain $\int cosh^2(u) du = \int \frac{1}{2} (cosh(2u) + 1)du = \frac{1}{4}sinh(2u) + \frac{1}{2} u + C$. Now use $sinh(2u) = 2 sinh(u)cosh(u)$ and we obtain:

$\frac{1}{2}sinh(u)cosh(u) + \frac{u}{2} + C$. The back substitution isn’t that hard if we recognize $cosh(u) = \sqrt{sinh^2(u) + 1}$ so we have $\frac{1}{2} sinh(u) \sqrt{sinh^2(u) + 1} + \frac{u}{2} + C$. Back substitution is now easy:

$\frac{1}{2} x \sqrt{x^2+1} + \frac{1}{2} ln(x + \sqrt{x^2 + 1}) + C$. No integration by parts is required and the dreaded $\int sec^3(x) dx$ integral is avoided. Ok, I was a bit loose about the domains here; we can make this valid for negative values of $x$ by using an absolute value with the $ln(x + \sqrt{x^2 + 1})$ term.