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 and Identities
- $ \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)
- $ \sinh^2 x + 1 = \cosh^2 x $
- $ \sinh^2 x = \frac{1}{2}(\cosh 2x - 1) $
- $ \cosh^2 x = \frac{1}{2}(\cosh 2x + 1) $
- $ \sinh 2x = 2\sinh x\cosh x $
- $ \cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x -1 = 2\sinh^2 x + 1 $
- $ \tanh \frac{x}{2} = \frac{\cosh x + 1}{\sinh x} $
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
Using:
- $ Ax = B\sinh \theta $
- $ dx = \frac{B}{A}\cosh \theta d\theta $
- $ \theta = \sinh^{-1} \frac{A}{B}x $
It can be shown that:
$ \int \frac{dx}{\sqrt{(Ax)^2+B^2}} = \frac{1}{A}\sinh^{-1} \frac{A}{B}x + C $
$ \int \sqrt{(Ax)^2+B^2}dx = \frac{\frac{A}{B}x\sqrt{(Ax)^2+B^2}+B\sinh^{-1}\frac{A}{B}x}{2}+C = \frac{\frac{A}{B}x\sqrt{(Ax)^2+B^2}+B\ln(\frac{A}{B}x+\frac{1}{B}\sqrt{(Ax)^2+B^2})}{2}+C $
Using:
- $ Ax = B\cosh \theta $
- $ dx = \frac{B}{A}\sinh \theta d\theta $
- $ \theta = \cosh^{-1} \frac{A}{B}x $
It can be shown that:
$ \int \frac{dx}{\sqrt{(Ax)^2-B^2}} = \frac{1}{A}\cosh^{-1} \frac{A}{B}x + C $
$ \int \sqrt{(Ax)^2-B^2}dx = \frac{\frac{A}{B}x\sqrt{(Ax)^2-B^2}-B\cosh^{-1}\frac{A}{B}x}{2}+C = \frac{\frac{A}{B}x\sqrt{(Ax)^2-B^2}-B\ln(\frac{A}{B}x+\frac{1}{B}\sqrt{(Ax)^2-B^2})}{2}+C $
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 $
