Difference between revisions of "Useful Formulas MA181Fall2008bell" - Rhea

Latest revision as of 09:12, 21 October 2008

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!

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)

$\sinh^2 x + 1 = \cosh^2 x$

$\sinh^2 x = \frac{1}{2}(\cosh 2x - 1)$

$\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)$

$\sinh 2x = 2\sinh x\cosh x$

$\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x -1 = 2\sinh^2 x + 1$

$\tanh \frac{x}{2} = \frac{\cosh x + 1}{\sinh x}$

--John Mason

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

$\frac{\pi}{4}= 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}$

$e^{ix} = \cos x + i \sin x$

Proofs of Euler's Formula

@@ Line 19: / Line 19: @@
 * <math>\sinh^2 x + 1 = \cosh^2 x</math>
-* <math>\sinh^2 x = \frac{1}{2}(\cosh 2x + 1)</math>
+* <math>\sinh^2 x = \frac{1}{2}(\cosh 2x - 1)</math>
-* <math>\cosh^2 x = \frac{1}{2}(\cosh 2x - 1)</math>
+* <math>\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)</math>
-* <math>\tanh \frac{x}{2} = \coth x + \text{csch} x</math>
+* <math>\sinh 2x = 2\sinh x\cosh x</math>
+* <math>\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x -1 = 2\sinh^2 x + 1</math>
+* <math>\tanh \frac{x}{2} = \frac{\cosh x + 1}{\sinh x}</math>
+--[[User:Jmason|John Mason]]
 == Trigonometric Integrations ==
@@ Line 33: / Line 39: @@
 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. --[[User:Jmason|John Mason]]
-== Hyperbolic Substitutions ==
+== [[Talk:Useful_Formulas_MA181Fall2008bell#Hyperbolic_Substitutions | 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.
+Using:
-<math> 1 + \sinh^2 \theta = \cosh^2 \theta </math>
+*<math> Ax = B\sinh \theta </math>
-which can lead to the identities:
+*<math> dx = \frac{B}{A}\cosh \theta d\theta</math>
-<math> \sqrt{1 + \sinh^2 \theta} = \cosh \theta </math>
+*<math> \theta = \sinh^{-1} \frac{A}{B}x </math>
-<math> \sqrt{\cosh^2 \theta - 1} = \sinh \theta </math>
+It can be shown that:
-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
+<math> \int \frac{dx}{\sqrt{(Ax)^2+B^2}} = \frac{1}{A}\sinh^{-1} \frac{A}{B}x + C </math>
-<math> \sinh^{-1} x = \int \frac{dx}{\sqrt{1+x^2}} = \ln (x + \sqrt{1+x^2}) </math>
+<math> \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 </math>
-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.
+Using:
-<math> x = \sinh \theta </math>
+*<math> Ax = B\cosh \theta </math>
-<math> dx = \cosh \theta </math>
+*<math> dx = \frac{B}{A}\sinh \theta d\theta</math>
-<math> \theta = \sinh^{-1} x </math>
+*<math> \theta = \cosh^{-1} \frac{A}{B}x </math>
-<math> \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 </math>
+It can be shown that:
-Expanding this to a general form:
-<math> \int \frac{dx}{\sqrt{(Ax)^2+B^2}} = \frac{1}{A}\sinh^{-1} \frac{A}{B}x + C </math>
-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:
 <math> \int \frac{dx}{\sqrt{(Ax)^2-B^2}} = \frac{1}{A}\cosh^{-1} \frac{A}{B}x + C </math>
-<s>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.</s>
+<math> \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 </math>
-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
-<math> \int \frac{dx}{\sqrt{1-x^2}} = 2\tan^{-1}\frac{\sqrt{1-x^2}}{1+x} + C </math>
-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.
---[[User:Jmason|John Mason]]
 ==Basic Integration Formulas==