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Math Help - integration

  1. #1
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    Question integration

    the integral from -1 to 2 of (cost +2tsint) dt

    Thank you very much.
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  2. #2
    MHF Contributor
    Joined
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    the integral from -1 to 2 of (cost +2tsint) dt

    The indefinite integral first:
    INT.[cost +2t*sint]dt
    = sint +INT.[2t*sint]dt

    Do the second integral by parts: INT.[u]dv = uv -INT.[v]du

    Let u = 2t, so, du = 2dt
    And dv = sint dt, so, v = -cost

    = sint +2t*(-cost) -INT.[-cost](2dt)
    = sint -2tcost +2sint +C
    = 3sint -2cost +C -------------(i) <--------edit: should be -2tcost.

    Then the definite integral:

    INT.(-1 --> 2)[cost +2t*sint]dt
    = [3sint -2cost] | (-1 --> 2)
    = [3sin(2) -2cos(2)] -[3sin(-1) -2cos(-1)]
    = [3sin(2) -2cos(2)] -[3sin(-1) -2cos(-1)]
    = 7.1652 -----------------------------------answer.


    -----------------------
    Edit: Sorry, but I forgot about the t in -2tcost.

    So let me correct the computations,

    Then the definite integral:

    INT.(-1 --> 2)[cost +2t*sint]dt
    = [3sint -2tcost] | (-1 --> 2)
    = [3sin(2) -2*2*cos(2)] -[3sin(-1) -2(-1)cos(-1)]
    = [3sin(2) -4cos(2)] -[3sin(-1) +2cos(-1)]
    = 5.836288 -----------------------------------answer.
    Last edited by ticbol; October 29th 2007 at 01:43 AM.
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