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Integrate 1/(x^2+y^2)^(3/2) 
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#1
Sep408, 09:03 PM

P: 6

Well, so I really want to integrate what's shown in the title:
i.e. [tex]\int \frac{dx}{(x^2+y^2)^\frac{3}{2}}[/tex] Now, I know there are quite a few straightforward answers to this. But what I really want is how people who do math got this formula in the first place. I don't just want a formula that seems to have come from a serendipitous accident or something. Please tell me how to derive the answer. (You might have guessed this has something to do with electric fields) Thank you for helping. 


#2
Sep408, 09:25 PM

P: 1,104

Hmm, are we keeping y constant here? If so then a trig substitution should might be a good first step.



#3
Sep408, 09:44 PM

P: 6

Yes, y is a constant.
Trig substitution? The final answer doesn't have any trig functions. 


#4
Sep408, 10:26 PM

P: 1,104

Integrate 1/(x^2+y^2)^(3/2)
That's because it probably involves an inverse trig function nested inside a trig function which of course would undo the trig function. Anyways you'll see what I mean if you try it.
Anyways I would try the substitution x = y*tan(t) (Have you learned usub?). 


#5
Sep508, 04:11 AM

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[itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so, dividing on both sides by [itex]cos(\theta)[/itex], [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex]. Thats why snipez90 suggested "tan(t)". Let [itex]x= ytan(\theta)[/itex]. In that case [itex]x^2+ y^2= y^2tan^2(\theta)+ y^2= y^2(tan^2(\theta)+ 1)= y^2sec^2(\theta)[/itex] so that [itex](x^2+ y^2)^{3/2}= (y^2sec^2(\theta))^{3/2}= y^3 sec^3(\theta)[/itex].
Also, it [itex]x= y tan(\theta)[/itex], then [itex]dx= ysec^2(\theta)d\theta[/itex]. So the integral, [itex]\int dx/(x^2+ y^2)^{3/2}[/itex] becomes [tex]\int \frac{y sec^2(\theta)d\theta}{y^3 sec^3(\theta)}= \int\frac{d\theta}{y^2 sec(\theta)}[/tex] [tex]= \frac{1}{y^2}\int cos(\theta)d\theta= \frac{1}{y^2}sin(\theta)+ C[/itex] Since [itex]tan(\theta)= x/y[/itex], imagine a right triangle having angle [itex]\theta[/itex], opposite side of length x, and near side of length y. Then the hypotenuse of that triangle has length [itex]\sqrt{x^2+ y^2}[/itex] and so [itex]cos(\theta)= y/\sqrt{x^2+ y^2}[/itex]. That means that [itex](1/y^2)cos(\theta)+ C= (1/y^2)(y/\sqrt{x^2+ y^2})+ C= 1/(y\sqrt{x^2+ y^2})+ C[/itex] 


#6
Sep508, 11:15 AM

P: 6

@snipez90:
Yes I've learned substitution. If it's solvable by substitution, then I just have the problem of finding what to substitute. @HallsofIvy: Ooh, that is one clever substitution. I never thought of that. What I don't get is why you are trying to find: [tex]{1 \over y^2} sin(\theta) + C = {1 \over y^2} {x \over \sqrt{x^2+y^2}} + C[/tex] Thanks so much! Freiddie 


#7
Sep508, 01:47 PM

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#8
Sep508, 03:40 PM

P: 6

(I digress  Is there a possibility that a Turing machine exists that calculates integrals?) 


#9
Sep608, 04:00 AM

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#10
Sep608, 06:23 AM

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If I were to define a function, say, f(x)= x^{5} 3x^{4}+ 3x^{3} 7 x^{2} 4x + 5, and ask "What is f(7)?", you would only need to evaluate itdo the arithmetic because you are already given the formula. But if I were ask "For what x is f(x)= 9" that would be a very difficult problem. It involves solving the equation and we know that, even for something as simple as a polynomial equation there may be several different solutions or no solution at all and, indeed, if the polynomial has degree 5 or more, there may be solutions that cannot be written in terms of algebraic operations. 


#11
Sep608, 10:52 AM

P: 6

Oh OK. Thanks for the explanation.



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