1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral of arcsinh(1/x)dx

  1. Nov 26, 2012 #1
    Hello, I am attempting to do a radiolgoical dose over equivalent fields problem. I have the following integral that I am trying to show each step for but am getting tricked up after the first integration. Any help would be greatly appreciated:

    1. The problem statement, all variables and given/known data
    ∫∫((1/√((x^2)+(y^2)))-μ')dxdy (μ' is a constant)also note the limits on X:0→L, and Y:0→W
    Solution: 2L*ln((D+W)/L)+2W*ln((L+D)/W)+μ'LW (where D=sqrt((W^2)+(L^2)))

    2. Relevant equations

    3. The attempt at a solution
    I can get through the first integration:
    Not sure what to do with the arcsinh, maybe express as logarithm?
  2. jcsd
  3. Nov 27, 2012 #2
    Yes, the inverse hyperbolic sine can be written as a logarithm: http://mathworld.wolfram.com/InverseHyperbolicSine.html . You should be able to write your first integral as ln(2(L + sqrt((L^2 + y^2))) - Lμ', though, which is different from what you got, although I may have read your integral badly. It doesn't get much nicer.
    The term sqrt(x^2 + y^2) may mean an easier integral in polar coordinates. The rectangular region of integration then changes to r in [0, Lsec(t)] and t in [0, arctan(W/L)] plus a second integral with r in [0, Wcsc(t)] and t in [arctan(W/L), pi/2]. Your integral then simplifies to
    [tex]\int_0^{\arctan\left(\frac{W}{L}\right)} \int_0^{L\sec\theta} 1 - \mu'r dr d\theta + \int_{\arctan\left(\frac{W}{L}\right)}^\frac{\pi}{2}\int_0^{W\csc\theta} 1 - \mu'r dr d\theta[/tex]
    This gives the same result with more transparent integrals.
  4. Nov 27, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Do you know how to convert to polar co-ordinates? The bounds are a bit tricky, but not too bad. For the second integral you'll need to integrate a secant (or two).
  5. Nov 27, 2012 #4
    Thank you for your help
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook