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Anyway to solve this integral?

  1. Aug 13, 2004 #1
    Int (x+ arcsin (x))/(sqrt(1-x^2))

    It's not working when I put U = (x+ arcsin (x)) and dU= 1/(sqrt(1-x^2)
    Is there anyway to solve it?

    Thanks
     
  2. jcsd
  3. Aug 13, 2004 #2
    If u = x + arcsin(x), then du = (1 + 1/sqrt(1 - x^2))dx. Maybe that's why it "isn't working".
     
  4. Aug 13, 2004 #3

    Galileo

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    Split the integral and remember (or look up) the derivative of arcsin(x)
     
  5. Aug 13, 2004 #4

    Zurtex

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    As Galileo says:

    [tex]\int \frac{x + \arcsin x}{\sqrt{1 + x^2}} dx = \int \frac{x}{\sqrt{1 + x^2}} dx + \int \frac{\arcsin x}{\sqrt{1 + x^2}}dx[/tex]

    Which makes the former integral of that quite easy, however the latter integral is of the form [itex]f(x) / f'(x)[/itex] which has no standard result. Although it does have a solution it is very long and extremely complex, are you sure this is the integral you were given?
     
  6. Aug 13, 2004 #5

    Galileo

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    The integrand is [tex]\frac{\arcsin x}{\sqrt{1 - x^2}}[/tex]
    with a minus sign, so the integrand has the form [itex]f(x)f'(x)[/itex] :smile:
     
  7. Aug 13, 2004 #6

    Zurtex

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    I'm really sorry, I don't know where my mind went, your totally right.
     
  8. Aug 13, 2004 #7
    Can I do it with UV - Integra (VdU)
     
  9. Aug 13, 2004 #8

    Zurtex

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    Your probably can but both the integrals are simple substiutions.
     
  10. Aug 13, 2004 #9
    can you explain more pls?
     
  11. Aug 13, 2004 #10

    Zurtex

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    Whenever you have an integral of the form:

    [tex]\int gf(x) * a * f'(x) dx[/tex]

    Where 'a' is some constant, then if you use the substitution [itex]u = f(x)[/itex] you get:

    [tex]du = f'(x)dx[/tex]

    [tex]a \int g(u) du[/tex]

    You can do this quite easily with both your integrals to get fairly easy integrals.
     
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