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Integrating cos^-2(x) after inverse substitution

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the substitution x=4sin(t) to evaluate the integral: S 1/[(16-x^2)^(3/2)] dx

    2. Relevant equations

    x = 4sin(t)

    3. The attempt at a solution

    x = 4sin(t)
    dx = 4cos(t) dt

    4cos(t) = (16-x^2)^(1/2), i cube both sides to get

    (4cos(t))^3 = (16-x^2)^(3/2), then plug in dx and denominator into the equation to get

    S 4cos(t)/[(4cos(t))^3] dt simplified and constant taken out i now get
    (1/16) S [cos^-2(t)] dt

    how do i integrate cos^-2(t) to get a simple answer. I don't think i can use the S cos^n(t) formula because then i have to integrate sin^-4(t) afterwards.

    any help would be awesome, thanks
  2. jcsd
  3. Feb 13, 2008 #2


    User Avatar

    cos^-2(x) = sec^2(x)

    Integral of sec^2(x) = tan(x)
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