Just in case so you don't have to look them up in your book or whatever. And so I can learn how to use Latex!
Contents
Hyperbolic Functions
- $ \sinh(x) = \frac{e^x - e^{-x}}{2} $
- $ \cosh(x) = \frac{e^x + e^{-x}}{2} $
- $ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $
- $ \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{{e^x + e^{-x}}}{{e^x - e^{-x}}} $
- $ \text{sech}(x) = \frac{1}{\cosh(x)} = \frac{2}{{e^x + e^{-x}}} $
- $ \text{csch}(x) = \frac{1}{\sinh(x)} = \frac{2}{e^x - e^{-x}} $
Idryg 20:10, 11 October 2008 (UTC)
Trigonometric Integrations
We know how to handle integrals of products of sines and cosines. If the number of sines and cosines are equal (i.e. their powers are equal), you can use this trick to drastically simplify the integral. It's a case that I don't think we covered in reading or lecture, but it did come up in a homework problem. Using the double-angle identity for sines:
$ \int\sin^n x \cos^n x dx = \int(\sin x \cos x )^n dx= \int (\frac{1}{2}\sin 2x)^n dx = \frac{1}{2^n} \int \sin^n 2x dx $
And then use the basic rules for powers of sines. It doesn't make the integrals definitively easier, but I like to simplify things down to single functions before integrating. You can't use this trick for different numbers of sines and cosines, unfortunately, as you end up with products of functions of different values. --John Mason
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 $
I suspect that 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. --John Mason
Basic Integration Formulas
$ \int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a} + C $
Other nifty formulas
$ \frac{\pi}{4}= 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} $
$ e^{ix} = \cos x + i \sin x $
