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Integration Technique

  1. Mar 24, 2007 #1

    I know [tex] \int \frac {1} {{\sqrt{1-x^2}} dx[/tex] is arcsine(x)

    My question is can [tex] \int \frac {1} {\sqrt{1-x^2}} dx[/tex]
    be solved by some technique (parts, substitution...) and the answer be in terms of x and NOT an expression of arcsine?

    Meaning, I would like the solution to the integral in terms of x and possibly other functions (ln for example) but not in terms of trig functions.

    If so, can you show me?

    Last edited: Mar 24, 2007
  2. jcsd
  3. Mar 24, 2007 #2
    It can be done by an appropriate trig substitution ie. x=sin(y), but the answer will end up identically. I'm fairly certain that arcsin(x) is the only answer you will ever get though.
  4. Mar 24, 2007 #3
    Antiderivatives are unique up to a constant. Ie, all antiderivatives of that will be arcsine(x)+C.
  5. Mar 24, 2007 #4


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    The only two other (sane) ways I can think of to write that antiderivative is as a power series, or in terms of a complex logarithm. But any of those ways will simply be a different way of writing arcsin x.
  6. Mar 24, 2007 #5
    One could certainly express arcsin in terms of logs using Euler's formula...
  7. Mar 24, 2007 #6

    Gib Z

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    Yea, Hurkyl already mentioned that.

    So yes I think the OP is happy,

  8. Mar 25, 2007 #7
    What I am wanting is an expression for arcsine that uses other functions (hopefully LN and powers of x) but I am hoping to work in the real domain.

    Gib Z help me quite a bit with a derivation regarding arcsine.

    I was hoping some clever integration could come up with another expression equal to arcsin by integrating the derivative of arcsine.

    Any thoughts?

  9. Mar 25, 2007 #8


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    If trig substitution, which just about everyone mentioned, is not a "clever integration", I don't know what is!
  10. Mar 25, 2007 #9
    I agree trig substitution is clever, I was hoping for a solution that doesn't include trig functions.

  11. Mar 26, 2007 #10

    Gib Z

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    I remember what sparky wanted from his previous thread that i tried to help him in. He had [tex]\arcsin x = -i \ln ( ix \pm \sqrt{1-x^2})[/tex] as he should, but could not prove why the expressions were equal.
    Last edited: Mar 26, 2007
  12. Mar 26, 2007 #11
    Correct Gib_Z and you did help me prove it - you helped me get the true solution and in that I ended up with some other solutions that I assumed involved the other quadrants.

    Using that expression (arcsin = -iln(...) has become very involved due to working in the complex domain.

    I'm now searching for a similar type of solution in the real domain only. (I suppose this rules out Euler's)

  13. Mar 28, 2007 #12

    Gib Z

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    Well then the best answer I can give you is the Taylor Series for Arcsin x, which If i remember correctly is this:

    [tex]\arcsin x = \sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}[/tex]
  14. Mar 28, 2007 #13
    Thank you all for your help.

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