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Integral of (1/2)sin2x dx

  1. Apr 28, 2015 #1
    The result of the integral of (1/2)sin2x dx with:

    upper limit x = arccos((R-h)/R)
    lower limit x = 0

    is (-h^2+2Rh)/(2R^2)

    I can not seem to get this exact answer my workings yield:

    let u = 2x, du/dx = 2 therefore dx = du/2

    Integral becomes (1/4) ∫ sinu du (with the same upper and lower limit)

    = -1/4 (cos(2x)) (evaluated at the limits)

    using cos2x = 2cos^2x-1 it becomes:

    = -1/4 (2cos^2x-1) (evaluated at the limits)

    = -1/4 ((2(cos*arccos ((R-h)/R))^2 - 1) - (2(cos*arccos(0))^2 -1)

    = -1/4 (2((R-h)/R)^2 - 1 + 1)

    = -1/2 ((R-h)/R)^2 = -1/2 - (h^2)/(2R^2) + h/R = (-h^2 + 2Rh - R^2)/(2R^2)

    As you can see, my answer has an extra R^2 term in the answer and I cannot see where I have gone wrong...
     
  2. jcsd
  3. Apr 28, 2015 #2

    SteamKing

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    Since your indefinite integral evaluates to -(1/4) * {2 cos2(x) - 1}, applying the limits of integration should be written as follows:

    I = -(1/4) * {2 cos2 (arccos ((R-h)/R)) - 1 - [2 cos2 (arccos (0)) - 1]}

    {Note: there is no '*' between the cos and the arccos. These are both functions which take arguments.}
     
  4. Apr 28, 2015 #3

    mfb

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    The last +1 in the second line should be -1. There is some bracket error in the first line I think.
     
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