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prove that the integral of arcsec(x) is equal to xarcsec(x) - ln|x+ sqrt((x^2)-1)

  • Calculus -

    Let x = secθ
    dx = secθtanθ dθ
    θ = arcsec x

    Int(arcsec(x) dx)
    = Int(θ secθtanθ dθ)

    now integrate by parts

    u = θ
    du = dθ

    dv = secθtanθ dθ
    v = secθ

    Int(θ secθtanθ) = θsecθ - Int(secθ dθ)

    To integrate secθ you have to resort to a trick:

    secθ (secθ + tanθ)/(secθ + tanθ) dθ

    now the top is sec^2θ + secθtanθ
    let u = secθ + tanθ and we have

    1/u du

    so, Int(secθ dθ) is ln|secθ + tanθ|
    and we end up with

    Int(θ secθtanθ) = θsecθ - ln|secInt(θ secθtanθ) = θsecθ - Int(secθ dθ) + tanInt(θ secθtanθ)
    = θsecθ - Int(secθ dθ)|

    Now, what does all that equal in x's?

    θ = arcsec(x)
    secθ = x
    tanθ = √(x^2-1)

    and you have your answer.

  • Calculus - PS -

    copy-paste error. That last complicated line should read:

    Int(θ secθtanθ) = θsecθ - Int(secθ dθ)
    = θsecθ - ln|secθ + tanθ|

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