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 1/sqrt(u^2-a^2) du, when u < -a

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Show and prove that where u < -a:

    integral of 1/(sqrt(u^2-a^2)) = ln(u+sqrt(u^2-a^2))+C or -cosh^-1(-u/a)+C

    2. Relevant equations
    when u > a, the
    integral of 1/(sqrt(u^2-a^2)) = ln(u+sqrt(u^2-a^2))+C or -cosh^-1(-u/a)+C

    My attempt at a solution: Don't I just use the same methodology as if u > a?

    integral 1/sqrt(-a^2+u^2) du
    For the integrand, 1/sqrt(u^2-a^2) substitute u = a sec(s) and du = a tan(s) sec(s) ds. Then sqrt(u^2-a^2) = sqrt(a^2 sec^2(s)-a^2) = a tan(s) and s = sec^(-1)(u/a):
    = integral sec(s) ds
    The integral of sec(s) is log(tan(s)+sec(s)):
    = ln(tan(s)+sec(s))+constant
    Substitute back for s = sec^(-1)(u/a):
    = ln((u (sqrt(1-a^2/u^2)+1))/a)+constant

    = ln(sqrt(u^2-a^2)+u)+constant
    Last edited: Feb 11, 2012
  2. jcsd
  3. Feb 11, 2012 #2


    User Avatar
    Science Advisor

    The difference is that the integral of 1/x is NOT ln(x), it is ln|x|. [itex]\int dx/x= ln|x|+ c[/itex]. Which sign you use for the absolute value will depend upon whether x> a or x< -a.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook