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Evaluate this integral

  1. Jan 20, 2008 #1
    1. The problem statement, all variables and given/known data

    elvaluate this integral (arcsinx)^2dx

    2. Relevant equations

    (arcsinx)^2dx

    3. The attempt at a solution

    integration by parts, let u= arcsinx and make y=arcsinx for easier integration. Once i plug it into the parts equation it turns into a mess. Any help would be superb.
     
  2. jcsd
  3. Jan 20, 2008 #2

    G01

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    This should be solvable by using substitution and integration by parts one after the other.

    HINT: Let sin u = x.
     
  4. Jan 20, 2008 #3
    That's what I've been doing and it still doesn't work out for me....oh well, I'll keep trying.
     
  5. Jan 20, 2008 #4
    [tex]\int(\sin^{-1}x)^2dx[/tex]

    [tex]t=\sin^{-1}x[/tex]
    [tex]dt=\frac{dx}{\sqrt{1-x^2}}[/tex]

    [tex]x=\sin t[/tex]
    [tex]dt=\frac{dx}{\sqrt{1-\sin^2 t}}[/tex]

    Continue simplifying and see what you can get.
     
    Last edited: Jan 20, 2008
  6. Jan 20, 2008 #5
    im getting xarcsinx+sqrt(1-x^2)
     
  7. Jan 20, 2008 #6
    Final answer? Take the derivative and see if you get your Integral.
     
  8. Jan 20, 2008 #7
    Im not sure how to do it your way so I just set it up by integration by parts. u=arcsinx, du=1/(sqrt(1-x^2)) dv=dx v=x. When I do it i get my answer above.
     
  9. Jan 20, 2008 #8
    After simplifying, the Integral becomes ...

    [tex]\int t^2 \cos t dt[/tex]
     
    Last edited: Jan 20, 2008
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