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Integrating Sec Tan^2

  1. May 25, 2009 #1

    TG3

    User Avatar

    The problem statement, all variables and given/known data
    Integrate the square root of (x^2 -1) over the interval 0-1.

    The attempt at a solution

    First off, I know this is a quarter of a circle, but I'm not supposed to solve it that way.
    Now then:

    x = sec theta
    dx= sec tan theta.

    square root (x^2-1) dx
    becomes
    square root (sec^2-1) sec tan

    Square root (tan^2) sec tan

    tan sec tan

    sec tan^2

    This is as far as I can get- I can't see any u substitutions that would work here, and don't know this integral off the top of my head either.
     
  2. jcsd
  3. May 26, 2009 #2
    You went from (tan^2)sec tan to just (tan^2)sec, it should be (tan^3)sec. That doesn't matter though, think about the graph and what the area is doing on the interval (0,1).
     
  4. May 26, 2009 #3
    Hogger: There was a square root.

    Integrals involving secants can be nasty; generally you have to integrate by parts. This is indeed the case with this integral, although first you should transform using identities to an integral only involving powers of secant. After integrating by parts once, you'll get an integral you've seen before, after which you can simply solve for it by algebraic means.
     
  5. May 26, 2009 #4

    Mark44

    Staff: Mentor

    The equation y = [itex]\sqrt{x^2 - 1}[/itex] is NOT the equation of a circle or any part of one.

    The function above results in imaginary numbers for x in the interval (-1, 1). Are you sure that you have copied the problem correctly?
     
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