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Homework Help: Spot the mistake! integration

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the integral of x * (arcsin x) * (1-x2)-1/2 dx


    2. Relevant equations
    integration by parts


    3. The attempt at a solution
    u = x, u' = 1
    v' = (arcsin x) * (1-x2)-1/2 (= f(x) * f'(x) )
    so v = ((arcsin x)2) / 2

    using integration by parts
    uv - integral of u'v

    = x * ((arcsin x)2) / 2 - integral of ((arcsin x)2) / 2


    Now use integration by parts for a second time to find the new integral, taking the half out as a constant:


    w = arcsin x w' = (1-x2)-1/2

    z' = arcsin x
    so z = x * (arcsin x) + (1-x2)1/2

    wz - integral of w'z

    = x * (arcsin x)2 + (arcsin x) * (1-x2)1/2 - integral of [
    x * (arcsin x) * (1-x2)-1/2 + 1]


    Substitute back into first equation (ie multiply above by -1/2)

    integral of x * (arcsin x) * (1-x2)-1/2 dx =

    x * ((arcsin x)2) / 2 - x * ((arcsin x)2) / 2 - ((arcsin x) * (1-x2)1/2) / 2 + 1/2 * [integral of x * (arcsin x) * (1-x2)-1/2 dx] + 1/2 * integral of 1 dx


    let [integral of x * (arcsin x) * (1-x2)-1/2 dx ] = I


    I = - ((arcsin x) * (1-x2)1/2) / 2 + I/2 + x/2

    I/2 = - ((arcsin x) * (1-x2)1/2) / 2 + x/2

    I = - ((arcsin x) * (1-x2)1/2) + x


    However when I check this by differentiation I end up with - x * (arcsin x) * (1-x2)-1/2. Hence I think the correct answer is ((arcsin x) * (1-x2)1/2) - x

    Thanks
     
  2. jcsd
  3. Sep 27, 2010 #2

    Mentallic

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    Homework Helper

    Ironically, the mistake was in your differentiation :tongue:
     
  4. Sep 27, 2010 #3
    Lol. Finally found where I was going wrong- differentiated (1-x2)1/2 aand forgot the minus sign. Thought I was going mad :P Thanks!
     
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