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Crazy Integral!

  1. Feb 19, 2007 #1

    Gib Z

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    Ok well The integral is :

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

    I can tell by inspection, it being of the form f'(x)/f(x), that the answers ln (arcsin x), but I was hopping Integration by parts could do it for me as well. But heres my Problem:

    u=1/(arcsin x)
    du=(-1)/ [sqrt(1-x^2) (arcsin x)^2] dx

    dv= 1/(sqrt(1-x^2) dx
    v= arcsin x

    uv- integral:v du

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

    The 2nd part is the Original integral. letting it equal I,

    I=1+I.

    What did i do wrong?
     
  2. jcsd
  3. Feb 19, 2007 #2

    morphism

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    This is what happens when you disregard the constant of integration. Your expression should actually be I + C_1 = 1 + I + C_2, where C_1 and C_2 are some constants.
     
  4. Feb 19, 2007 #3

    Gib Z

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    Ahh Thats what came to mind after I posted, but then i thought the constants would cancel >.<", well how do we do the integral then?
     
  5. Feb 19, 2007 #4

    morphism

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    Not by parts.
     
  6. Feb 19, 2007 #5

    cristo

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    Try the substitution u=arcsinx..
     
  7. Feb 19, 2007 #6

    Gib Z

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    >.<" Ok..Im feeling really stupid right now...Someone please shoot me.
     
  8. Feb 19, 2007 #7

    morphism

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    Which is what he did here:
     
  9. Feb 19, 2007 #8
    I don't think that that is correct as stated. It looks more like 1/[f'(x)*f(x)] to me.
     
  10. Feb 20, 2007 #9

    dextercioby

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    No, no, he's right. The derivative is in the numerator.

    This integral is one good example of the situation when some integrals have a unique method of solving.
     
    Last edited: Feb 20, 2007
  11. Feb 20, 2007 #10
    Yes, of course, you're right, I don't know what I was thinking.
     
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