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Trig substitution into integrals

  1. Feb 15, 2014 #1
    I was testing for convergence of a series:
    ∑[itex]\frac{1}{n^2 -1}[/itex] from n=3 to infinity

    I used the integral test, substituting n as 2sin(u)

    so here's the question:
    when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine.

    Is it still possible to make the substitution? Or is there a restriction when this happens?
     
  2. jcsd
  3. Feb 15, 2014 #2

    Mark44

    Staff: Mentor

    What does "fit inside the sine" mean?
    Sure, you can make the substitution. The integral will be from 3 to, say b, and you take the limit as b → ∞.

    Not that you asked, but it's probably simpler and quicker to break up 1/(n2 - 1) using partial fractions.
     
  4. Feb 16, 2014 #3
    Inside the sine meaning, the argument of the 'arcsine' would only range from -1 to 1.
    So I'm guessing you can't make the substitution because arcsin(infinity) = error?
     
  5. Feb 16, 2014 #4
    If you're looking for an appropriate trig substitution for the definite integral (and not just one that gets you a correct antidierivative), then ##\sec u## is the way to go. But like Mark44 said, partial fractions is really the "right" technique of integration for this particular integral.
     
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