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Homework Help: Integral of ArcCos Problem

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    The region between y=cos x and the x-axis for x [tex]\in[/tex] [0, [tex]\pi[/tex]/2] is divided into two subregions of equal area by a line y=c. Find c.

    2. The attempt at a solution
    First I drew a graph of the region bounded between the function and the x-axis in [0, [tex]\pi[/tex]/2]. Next I found the total area of the region:
    [tex]A_{1,2}[/tex] = [tex]\int^{\pi/2}_{0}[/tex] cos x dx = sin x |[tex]^{\pi/2}_{0}[/tex] = 1

    Dividing the total area by 2, gives the area of each portion divided by y=c.

    [tex]A_{1}[/tex] = 1/2 = [tex]\int^{c}_{0}[/tex] [tex]cos^{-1}[/tex] (y) dy = y[tex]cos^{-1}[/tex] (y) - [tex]\sqrt{1-y^{2}}[/tex] [tex]|^{c}_{0}[/tex]
    = c (cos[tex]^{-1}[/tex] (c)) - [tex]\sqrt{1-c^{2}}[/tex] + 1

    --> c ([tex]cos^{-1}[/tex] (c)) - [tex]\sqrt{1-c^{2}}[/tex] = -1/2

    I'm not sure how to solve for c from the equation above. Any input will be appreciated.
    Last edited: Jan 31, 2009
  2. jcsd
  3. Jan 31, 2009 #2


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    Homework Helper
    Gold Member

    I don't think think an exact solution is possible. Try estimating the solution using Newtons Method or any other numerical technique you have learned.
  4. Jan 31, 2009 #3


    Staff: Mentor

    You're not going to be able to solve for an exact value of c using algebraic techniques. The best you can do is to get an approximate value. A simple way to do this is to start with an educated guess, like, say c = 0.4. (The value has to be less than 1/2 because there is more area in the lower part of the region than in the upper part.)

    You want to find a value c so that c*cos^(-1)(c) = sqrt(1 - c^2) + 1/2 = 0. c = .4 probably isn't right, but it will give you an idea of what to use for your next try. Keep doing this, with better and better approximations until your successive approximations are in agreement in two, three, or four decimal places, or however precise you want to be.
  5. Jan 31, 2009 #4
    I had thought it would be just approximating. Thank you =)
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