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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!
     
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