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Math Help - Some calc questions

  1. #1
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    Some calc questions

    I'm having problems with some definite intergral questions

    Int2-1 (Ln(x)/x^2) dx
    I got 1 as a result from that, but it's different from the answer, I'm wondering about how to do it?

    Also
    Int cos(Ln(x)) dx


    Can anyone also tell me what the derivative of cos^-1, tan^-1, sin^-1.. etc
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  2. #2
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    Quote Originally Posted by nirva
    I'm having problems with some definite intergral questions

    Int2-1 (Ln(x)/x^2) dx
    I got 1 as a result from that, but it's different from the answer, I'm wondering about how to do it?

    Also
    Int cos(Ln(x)) dx


    Can anyone also tell me what the derivative of cos^-1, tan^-1, sin^-1.. etc
    1) INT.(2->1)[lnX / X^2]dX --------------(1)
    By parts.
    INT.[u]dv = uv -INT.[v]du ------(i)

    Let u = lnX
    so, du = (1/X)dX

    And dv = dX / X^2
    So, v = -1/X

    Then, for the indefinite integration of (1),
    INT.[lnX / X^2]dX
    = (lnX)(-1/X) -INT.[-1/X](dX /X)
    = -(lnX)/X +INT.[1/X^2]dx
    = -(lnX)/X +(-1/X) +C
    = -(LnX)/X -1/X +C
    And so for the definite integration of (1),
    = -[(lnX)/X +1/X](2->1)
    = -[ln(1)/1 +1/1] +[ln(2)/2 +1/2]
    = -1 +(1/2)ln(2) +1/2
    = (1/2)[ln(2) - 1] -----------answer.

    ------------------------------------------
    2) INT.[cos(lnX)]dx -----(2)

    Whoa, this could be long, so let me skip some steps.

    Let U = lnX
    So, X = e^U

    and, dU = dX /X
    dX = X*dU = e^U dU

    Then, substitute those into (2),
    INT.[cos(lnX)]dX
    = INT.[cosU](e^U dU) ------(2a)

    Integrate (2a) by parts:
    INT.[r]ds = r*s -INT.[s]dr
    r = e^U; so, dr = e^U dU
    ds = cosU dU; so, s = sinU
    Then, back to (2a),
    = e^U *sinU -INT.[sinU](e^U dU)
    = e^U *sinU -INT.[e^U](sinU dU) -------(2b)

    Integrate the last part of (2b) by parts again,
    INT.[m]dn = m*n -INT.[n]dm
    m = e^U; so, dm = e^U dU
    dn = sinU dU; so, n = -cosU
    Then, back to (2b),
    = e^U *sinU -{e^U *(-cosU) -INT.[-cosU](e^U dU)}
    = e^U *sinU +e^U *cosU -INT.[cosU](e^U dU) -------(2c)

    The last part of (2c) is the same as (2a),
    And since (2a) is (2) also, then,
    = e^U *sinU +e^U *cosU -INT.[cos(lnX)]dX

    Transpose the last part to the unseen LHS,
    2(INT.[cos(lnX)]dX) = e^U *sinU +e^U *cosU
    So, divide both sides by 2,
    INT.[cos(lnX)]dX = (1/2)[e^U *sinU +e^U *cosU]
    Substituting back away from U,
    INT.[cos(lnX)]dX = (1/2)[X*sin(lnX) +X*cos(lnX)]

    Therefore,
    INT.[cos(lnX)]dX = (X/2)[sin(lnX) +cos(lnX)] +C. --------answer

    -------
    For general derivatives or integrals, go to this website:
    http://www.sosmath.com/tables/tables.html
    (Sorry, I do not know how to link.)

    --------------
    Hey, it automatically linked. Amazing.
    Last edited by ticbol; May 13th 2006 at 11:24 PM. Reason: Integral Table
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