
#1
Dec412, 03:25 PM

P: 107

As the title obviously states, how can I evaluate this integral by hand ? I know the result of it, I need to learn how to do it.
∫(z^{2}+x^{2})^{3/2}dx Thanks in advance. 



#2
Dec412, 03:29 PM

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Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.




#3
Dec412, 03:57 PM

P: 107





#4
Dec412, 04:05 PM

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How to compute this integral ?it is NOT a particularly easy integral to evaluate, in that it is rather lengthy to do so. That is probably why your book skipped it. 



#5
Dec412, 04:08 PM

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x=z tan t can also do the trick.




#6
Dec412, 04:11 PM

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Scribbling out the solution with tan(t) or Sinh(t) takes about the same amount of time and space, though.. 



#7
Dec512, 09:03 AM

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The actual derivation is rather lengthy, but here it is:
1. x=zSinh(t). Thus, we have: [tex]dx=z\cosh(t)dt[/tex] [tex]z^{2}+x^{2}=z^{2}(1+\sinh^{2}(t))=z^{2}\cosh^{2}(t)[/tex] Thus, the integral can be simplified to: [tex]\int\frac{dt}{zz\cosh^{2}(t)}[/tex] 2. This is readily integrated to: [tex]\frac{\tanh(t)}{zz}=\frac{x}{z^{2}\sqrt{z^{2}+x^{2}}}[/tex] 



#8
Dec612, 09:25 AM

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Arildno really likes the hyperbolic function substitutions. Personally, I prefer trig substitutions, perhaps only because they were the first ones I learned.
We know, of course, that [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] and, dividing through by [itex]cos^2(\theta)[/itex], [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex]. So if we let [itex]x= z tan(\theta)[/itex], [itex]z^2+ x^2= z^2+ z^2tan^2(\theta)= z^2(1+ tan^2(\theta)_= z^2sec^2(\theta)[/itex]. Of course, [itex]dx= z sec^2(\theta)d\theta[/itex] so the integral becomes [tex]\int\frac{z sec^2(\theta)}{z^3 sec^3(\theta)}d\theta= \int \frac{1}{sec(\theta)}d\theta[/tex] [tex]= \int cos(\theta) d\theta[/tex] which is easy. Since [itex]\theta= arctan(x/z)[/itex], the integral will eventually give [itex]sin(arctan(x/z))[/itex]. You can imagine that as describing a right triangle with legs x and z (x opposite the angle) so that the hypotenuse has length [itex]\sqrt{x^2+ z^2}[/itex] and [itex]sin(arctan(x/z))= \frac{x}{\sqrt{x^2+ z^2}}[/itex]. 



#9
Dec612, 02:13 PM

P: 9

Sorry that this is a bit late, but is Z a real number?
If not then you could use Cauchy: nonholomorphic points will be at z^2 =  x^2 ( e.g. x =1, z = i). 


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