- Hey Ian, I like the work. As a hint though, trig and hyperbolic trig functions can be made to look nicer by using a \ before them. (This works because \cos, \sin, \tan, etc are symbols in latex, i think.
- Great Work! If you want more syntax tricks, look at this table of formula. --Mboutin 11:41, 17 October 2008 (UTC)
Hyperbolic Substitutions
With all the similarities between the trigonometric and hyperbolic functions, it occurred to me that the latter may be as useful in substituting in integrals as the former. After some minor manipulation and some basic algebra to prove it, I quickly found that a relationship similar to the Pythagorean trigonometric identity existed for the hyperbolic sine and cosine.
$ 1 + \sinh^2 \theta = \cosh^2 \theta $
which can lead to the identities:
$ \sqrt{1 + \sinh^2 \theta} = \cosh \theta $
$ \sqrt{\cosh^2 \theta - 1} = \sinh \theta $
You may notice that these are the forms for which we previously substituted tangent and secant, respectively. To be honest, those two methods work out with extremely messy solutions. As Prof. Bell easily showed, you can use tangent substitution to prove that
$ \sinh^{-1} x = \int \frac{dx}{\sqrt{1+x^2}} = \ln (x + \sqrt{1+x^2}) $
which is a really impressive fact, but is a horribly frightening function. We can use hyperbolic sine substitution to directly show that integral is equal to inverse hyperbolic sine (plus a constant); not an incredible feat, but it does show the validity of the method. In fact, it's very simple.
$ x = \sinh \theta $
$ dx = \cosh \theta $
$ \theta = \sinh^{-1} x $
$ \int \frac{dx}{\sqrt{1+x^2}} = \int\frac{\cosh \theta}{\sqrt{1+\sinh^2 \theta}} d\theta = \int\frac{\cosh \theta}{\cosh \theta} d\theta = \int d\theta = \theta + C = \sinh^{-1} x + C $
Expanding this to a general form:
$ \int \frac{dx}{\sqrt{(Ax)^2+B^2}} = \frac{1}{A}\sinh^{-1} \frac{A}{B}x + C $
I suspect that this isn't really a fair way to represent the integral on classwork, but if you are computing something, it is certainly easier on the eyes (and the brain!) to write this form. If you can remember the definition of the inverse hyperbolic sine in the terms of logs and square roots, then you can really cheat the system. This works just as well for hyperbolic cosine:
$ \int \frac{dx}{\sqrt{(Ax)^2-B^2}} = \frac{1}{A}\cosh^{-1} \frac{A}{B}x + C $
Using hyperbolic tangent or secant may allow you simplify the other form (the one we usually use sine for). In that case, sine may still be better option. I'm still playing around with the hyperbolics.
That turns out to be very much not the case. Using a table of integrals, a definition of the inverse hyperbolic tangent, and some identities which I had to play around with to find, it can be shown that
$ \int \frac{dx}{\sqrt{1-x^2}} = 2\tan^{-1}\frac{\sqrt{1-x^2}}{1+x} + C $
I'm not even sure that is correct.
Substituting with hyperbolic secant gives an even worse answer. So much for that. I've yet to find any other function that is easier to integrate with hyperbolic substitutions; let me know if you find one. So far, all I have really done is shown some general methods for using the hyperbolics to find the integrals of their inverses.
I have found two more cases where you can use this substitution. In class, Prof. Bell solved
$ \int\sqrt{4x^2+1}dx $
The solution took so long and so many steps that I stopped taking notes a few steps before he finished. To solve it, he substituted tangent and ended up with an integral of secant cubed, which was a mess to integrate. With hyperbolic sine in its place, the integral then becomes the integral of hyperbolic cosine squared, which is not only a lower power, but can easily be integrated by stating the hyperbolic cosine in terms of the exponential function. You can then reverse the substitution for a solution, which happens to be (I think) the same solution we found in class. Also, it solves in terms of the inverse hyperbolic sine, simplifying the solution drastically. The same method applies to the integral of a square root of the variable squared minus a constant, except you instead substitute hyperbolic cosine.
I have added these integrals to this section, in terms of A's and B's. It seems that this method is best applied in situations where normal trigonometric substitution leads to awkward integration. I have yet to find a situation where substituting hyperbolic secant or tangent is helpful.
