1. Not finding help here? Sign up for a free 30min 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!

Integration- hyperbolic trig can't find my mistake

  1. Jan 6, 2008 #1
    1. The problem statement, all variables and given/known data
    (I haven't got the thing to write equations so this is a little difficult to understand sorry!)
    Show that the integral of (4x^2 - 1)^-1/2 from 0.625 to 1.3 is equal to 1/2(ln 5 - ln 2)

    2. Relevant equations
    my formula book states:
    the integral of (x^2 - a^2)^-1/2 is arcosh (x/a) or ln (x + (x^2 - a^2)^1/2)

    3. The attempt at a solution
    the original integral equals:
    1/2 the integral of (x^2 - 1/4)^-1/2
    comparing this to the stated result I get a^2 = 1/4
    so substituting in limits:
    = 1/2 ln(1.3 + (1.3^2 -1/4)^1/2) - 1/2 ln(0.625 + (0.625^2 - 1/4)^1/2)
    =1/2 (ln 2.5 - ln 1)

    which is incorrect. I just about follow the worked solution, which uses a very slightly different method, but I want to know why this is incorrect because I can't find a mistake in it.. Thanks!
    Last edited: Jan 6, 2008
  2. jcsd
  3. Jan 6, 2008 #2
    It is correct

    [tex]\frac 1 2(\ln{\frac 5 2}-\ln 1) =\frac 1 2(\ln 5 - \ln 2)[/tex]
  4. Jan 6, 2008 #3
    oh yeah.. oops! I didn't realise.
    Thank you!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integration- hyperbolic trig can't find my mistake