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Integration using Hyperbolic Trig substitution

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate:

    [tex]\int\\{1}/{\sqrt{x^2-1}}[/tex] dx between -3, -2

    I know I'm supposed to use hyperbolic substitution in the question.

    2. Relevant equations

    edit: cosh^2(t) - sinh^2(t) = 1

    3. The attempt at a solution

    let x = -cosht, inside the integral let dx = sinh(t) dt

    int ( sinht / sqrt(cosh^2t - 1) ) dt

    now normally at this point i would use the trig identity cosh^2(t) + sinh^2(t) = 1 to eliminate the -1 in the root, however i can't work out how to sub in to eliminate because i've had to use a negative cosht for x as the bounds of the integral are both negative.

    I'm not great at this area of maths, I thought maybe if i substituted sinh(t) for x, I could possible rewrite -sinh(t) as sinh(-t) but i'm not sure if that's a fair solution or not.
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    rock.freak667

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    your identity is incorrect, it's supposed to be cosh2t-sinh2t=1
     
  4. Mar 15, 2009 #3
    cheers.

    that still means i can't sub cosht for x but i can use sinh = x

    then everything cancels and im left with int dt
    which is -sinh^-1(x)

    which gives me:

    (1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995

    still not sure if this is technically accurate
     
  5. Mar 15, 2009 #4
    this is the current solution im working on:

    dx = cosht dt
    x = -sinht

    therefore;

    1/ sqrt(x^2 - 1 ) = cosht dt / sqrt( -sinh^2 t - 1 ) ----- (1)
    using the trig id:
    cosh^2 t - sinh^2 t = 1
    -sinh^2 t = -cosh^2 t + 1 ----- (2)

    (2) into (1)

    int (cosht / sqrt(-cosh^2 t)) dt = int dt
    int dt = t + c
    t = 1/(-sinh x)

    therefore the integral evaluated is:
    (1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995
     
    Last edited: Mar 15, 2009
  6. Mar 15, 2009 #5

    rock.freak667

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    so if you have [itex]\int dt =t +c[/itex]

    and x=cosh(t), doesn't it mean that t=cosh-1(x) ?
     
  7. Mar 15, 2009 #6
    Sorry, x = -sinht, because the domain of the integral is negative.

    i switched around the dx and x substitutions the second time around.
     
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