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Integral [ArcSin(x)]dx

  1. May 6, 2007 #1
    Hi, I'm from in Brasil and I need some help..
    I have no sucess resolving Integral [ArcSin(x)]dx ..

    Using Integration by parts, i don't kno what to do in expression 'c':

    Integral [ArcSin(x)]dx = x.ArcSin(x) - Integral[x.(1 - x^2)^-1/2]
    ---------a---------- -----b----- ------------c-------------

    How expression 'c' turns into only (1 - x^2)^1/2 ?

    Thanks for all.

    PS: excuse me for possible [a lot of] gramathical erros .. i don't speak english..:uhh:
  2. jcsd
  3. May 6, 2007 #2


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    [tex]I= \int \sin^{-1}(x)dx = x \sin^{-1}(x) - \int x (1-x^2)^{-1/2}dx[/tex]
    Now use trig substitution:
    [tex] x = \sin(t), dx = cos(t)dt[/tex]
    [tex] I = x\sin^{-1}(x) - \int \sin(t)\cos(t)(1-\sin^2(t))^{-1/2}dt=\int \sin(t) \frac{\cos(t)}{\cos(t)}dt[/tex]
    [tex] = x\sin^{-1}(x) - \int \sin(t)dt = x\sin^{-1}(x)-\cos(t)+C[/tex]***
    [tex] = x\sin^{-1}(x) - \cos(\sin^{-1}(x)) + C[/tex]
    [tex] = x\sin^{-1}(x) - \sqrt{1 - x^2} + C[/tex]

    Or you can use a substitution immediately:
    [tex]\int \sin^{-1}(x)dx = \int t \cos(t)dt[/tex]
    [tex] x = sin(t)[/tex]
    and then integrate by parts:
    [tex] = \int t cos(t)dt= t\sin(t) - \int \sin(t)dt[/tex]
    [tex] = \sin^{-1}(x) x - \int \sin(t)dt=x\sin^{-1}(x)-\cos(t)+C[/tex]
    which is the same as (***) in the previous try.
    Last edited: May 6, 2007
  4. May 6, 2007 #3


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    You could also use the much easier substitution [tex] t=x^2 [/tex] to evaluate [tex] \int x (1-x^2)^{-1/2}dx[/tex] :smile:
    Last edited: May 6, 2007
  5. May 6, 2007 #4


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    And even better, you can try the substitution: t = 1 - x2. :)
  6. May 6, 2007 #5
    All Right!

    One more time, thanks for all.

    VietDao29, I take your sugestion [so many simple and usefull] and
    it works fine. Thank you.

    See you..
  7. Oct 8, 2009 #6
    Can someone explain this part please?
  8. Oct 8, 2009 #7


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    Dearly Missed

    Use integration by parts, noting that:
  9. Oct 8, 2009 #8
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