# Integral of (x^2+y^2)^(-3/2)dy

1. Sep 15, 2014

### FlyingButtress

1. The problem statement, all variables and given/known data

Hello, I came across an unusual integral problem in my Physics book that I could not solve. The book simply skips the work and says to check the integral tables, but I wasn't able to find one that satisfied this problem.

2. Relevant equations

This is the problem: integral of (x^2+y^2)^(-3/2)dy

3. The attempt at a solution

Tried u-sub, but the lack of y makes it impossible. Couldn't do trig sub because of the negative exponent.

When I plugged it into WolframAlpha, it gave me a tidy solution, but the solutions are unavailable even in their app. It loads for a bit and shows "solution unavailable."

The proper solution should be y/[(x^2)*√(x^2+y^2)]+constant.

Here is the actual URL:
http://www.wolframalpha.com/input/?i=integral+of+(x^2+y^2)^(-3/2)dy&incTime=true

This is the second time coming across this integral in this chapter and I cannot solve it for the life of me. Any insights would be greatly appreciated. Thanks!

2. Sep 15, 2014

### dirk_mec1

The x can be regarded as a constant, right? Is it part of a double integral? Can you use IBP?

Last edited: Sep 15, 2014
3. Sep 15, 2014

### FlyingButtress

Yes, x could be regarded as a constant, and no, not a part of a double integral. The physics parts of the problem were constants, so I just uploaded the core variables.

IBP didn't work for me because when I used tabular integration because the derivative and integral of each part would continue indefinitely if that makes sense.

Also if it helps, I worked through my problem using this guide I found:
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html [Broken]

I only switched the x and y because my problem was a horizontal version of this example.

Last edited by a moderator: May 6, 2017
4. Sep 15, 2014

### B3NR4Y

I'm not sure what you mean by trig not working due to the negative exponent. I took a shot at it and got the same result as wolfram alpha. Try to take $y = x tan \theta$. so the integral becomes:

$\int \frac{1}{(x^{2} + y^{2})^{\frac{3}{2}}} dx = \int \frac{sec^{2} \theta d\theta}{(x^{2} + x^{2}tan^{2} \theta)^{\frac{3}{2}}}$

With trigonometric identities the solution should be easy.

Edit: wooah that LaTeX screw up, going to fix it.

5. Sep 15, 2014

### dirk_mec1

Yes trigoniometrics is the way to go here.

6. Sep 15, 2014

### FlyingButtress

Wow, that was easy. I dunno how that didn't spring to mind; I really need to brush up on my math haha.

Thank you!!

7. Sep 15, 2014

### Ray Vickson

What do you mean that in Wolfram it gives a tidy solution, but the solutions are unavailable? When I click on your link it gives a solution.

8. Sep 15, 2014

### FlyingButtress

Oops, I left out a word there. Meant to say "solution steps." Usually in their app it shows the procedure of obtaining the solution. By "tidy" I meant that the end solution looked elegant and nothing like what I had on my paper at the time...