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

  • #1

Homework Statement


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

Homework 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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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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.
 

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