integrate 1/(x^2+y^2)^(3/2)


by Freiddie
Tags: 1 or x2, integrate, y23 or 2
Freiddie
Freiddie is offline
#1
Sep4-08, 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.
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
snipez90
snipez90 is offline
#2
Sep4-08, 09:25 PM
P: 1,106
Hmm, are we keeping y constant here? If so then a trig substitution should might be a good first step.
Freiddie
Freiddie is offline
#3
Sep4-08, 09:44 PM
P: 6
Yes, y is a constant.

Trig substitution? The final answer doesn't have any trig functions.

snipez90
snipez90 is offline
#4
Sep4-08, 10:26 PM
P: 1,106

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 u-sub?).
HallsofIvy
HallsofIvy is offline
#5
Sep5-08, 04:11 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877
[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]
Freiddie
Freiddie is offline
#6
Sep5-08, 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:

Quote Quote by HallsofIvy View Post
[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]
when there's a sin(x) in the final equation. In that case, the answer would be:
[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
ObsessiveMathsFreak
ObsessiveMathsFreak is offline
#7
Sep5-08, 01:47 PM
P: 406
Quote Quote by Freiddie View Post
I don't just want a formula that seems to have come from a serendipitous accident or something.
To put it to you bluntly, this is in fact where most integration formulae come from. Things like the above proof usually only come after the discovery of the initial function.
Freiddie
Freiddie is offline
#8
Sep5-08, 03:40 PM
P: 6
Quote Quote by ObsessiveMathsFreak View Post
To put it to you bluntly, this is in fact where most integration formulae come from. Things like the above proof usually only come after the discovery of the initial function.
That's a depressing thought.

(I digress - Is there a possibility that a Turing machine exists that calculates integrals?)
Ben Niehoff
Ben Niehoff is offline
#9
Sep6-08, 04:00 AM
Sci Advisor
P: 1,562
Quote Quote by Freiddie View Post
(I digress - Is there a possibility that a Turing machine exists that calculates integrals?)
No, there is no algorithm for finding general integrals (contrast to derivatives, for which there are very clear-cut algorithms). Many integral formulas were found, as mentioned, by lucky guesses (although many of those are still arrivable deterministically, if one is clever enough).
HallsofIvy
HallsofIvy is offline
#10
Sep6-08, 06:23 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877
Quote Quote by Ben Niehoff View Post
No, there is no algorithm for finding general integrals (contrast to derivatives, for which there are very clear-cut algorithms). Many integral formulas were found, as mentioned, by lucky guesses (although many of those are still arrivable deterministically, if one is clever enough).
In fact, this is generally true of "inverse" problems.

If I were to define a function, say, f(x)= x5- 3x4+ 3x3- 7 x2- 4x + 5, and ask "What is f(7)?", you would only need to evaluate it-do 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.
Freiddie
Freiddie is offline
#11
Sep6-08, 10:52 AM
P: 6
Oh OK. Thanks for the explanation.


Register to reply