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Trig substitution

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    consider the region bounded by the graphs of y=arcsinx, y=0, x = 1/2.

    a) find the area of the region.
    b) find the centroid of the region.

    2. Relevant equations

    [itex]\displaystyle\int_0^{1/2} {arcsinx dx}[/itex]

    u=arcsinx; du = [itex]\frac{1}{1-x^2}dx[/itex]

    dv=dx ; v=x

    xarcsinx][itex]^{1/2}_{0}[/itex] - [itex]\displaystyle\int_0^{1/2} {\frac{x}{\sqrt{1=x^2}} dx}[/itex]

    = (1/2arcsin(1/2) - 0) + [itex]\sqrt{1-x^2}^{1/2}_{0}[/itex]

    answer: [itex]\frac{\pi + 6\sqrt{3}-12}{12}[/itex]

    3. The attempt at a solution

    i can't get that answer.

    in my work i have:

    xarcsix][itex]^{1/2}_{0}[/itex] = [itex]\frac{\pi}{12}[/itex]

    for the other part i get:

    [itex]\sqrt{1-x^2}^{1/2}_{0}[/itex] = [itex]\sqrt{1-\frac{1}{4}} - 1[/itex]

    which i can see how the last term -1 will go to -12/12, and pi/12 goes if front, but how are the getting the middle term, 6[itex]\sqrt{3}[/itex]?
    Last edited: Mar 12, 2013
  2. jcsd
  3. Mar 12, 2013 #2


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    It seems to me that this would be much easier to do by looking at x as a function of y.
  4. Mar 12, 2013 #3
    I think for this problem, it's just easier to do a substitution at the beginning, x=sinθ. Then you just do a integration by parts.

    Note this only works because the limits of integration are between 0 and 1/2. Also, arcsin(sinx)=x.
  5. Mar 12, 2013 #4


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    The middle "term" you want to get is [itex]\frac{6\sqrt{3}}{12}[/itex], not [itex]6\sqrt{3}[/itex].

    From [itex]\sqrt{1-\frac{1}{4}}[/itex], do the subtraction, and split into two square roots, one in the numerator, and one in the denominator. It's straightforward from there.
  6. Mar 12, 2013 #5
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