- 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.
