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Math Help - integration

  1. #1
    Newbie
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    May 2009
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    integration

    integrate 1/ (y^2 - 1) with respect to dy.
    no idea how.. ;0
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  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
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    There are 2 ways to go here

    1. Use partial fraction decomposition

    1/(y^2-1) = 1/(y-1)(y+1) = A/(y-1) + B/(y+1)

    1 = A(y+1) + B (y-1)

    if y= 1 A =1/2 If y = -1 B= -1/2

    1/(y^-1) = 1/2[1/(y-1) - 1/(y+1)]

    you have 2 simple integrals

    2. Trig sub

    y =sec(t) dy = sec(t)tan(t) y^2 - 1 = tan(t)

    You have int[ sec(t)/tan(t)] = int [ csc(t)] = -ln|csc(t) +cot(t)| +c

    convert back to fn of y by drawing right triangle with t = arcsec(y)

    so you're triangle has hypotneuse y adjacent side 1 opposite side

    (y^2 -1)^(1/2)
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