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The integration of arcsin(x)

  1. Dec 29, 2009 #1
    Hello all,

    I've solved this equation to get:

    [tex]
    \int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C
    [/tex]

    using integration by parts. I have found, however, that my textbook has
    the part

    [tex]-\sqrt{1-x^2}[/tex]

    instead of a +, leaving the answer

    [tex]
    \int \arcsin{(x)}dx = -\sqrt{1-x^2}+ x\arcsin{(x}) + C
    [/tex]

    This minus sign has been confirmed by a website I came across
    Code (Text):
    [PLAIN]http://math2.org/math/integrals/tableof.htm
    [/PLAIN] [Broken]
    Elsewhere on this forum I've seen the answer with the plus sign:
    Code (Text):
    https://www.physicsforums.com/showthread.php?t=89216
    giving rise to this
    inconsistency.

    Any help is appreciated and thanks in advance,
    Charismaztex
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 29, 2009 #2

    dextercioby

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    It's with the plus sign. (EDIT: See the 2 posts below from <arildno> and me).
     
    Last edited: Dec 29, 2009
  4. Dec 29, 2009 #3
    Yeah, that's what I calculated but my calculator and textbook says otherwise...
     
  5. Dec 29, 2009 #4

    arildno

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    Dearly Missed

    It depends on which interval arcsine maps [-1,1] onto.

    Conventionally, arcsine maps this interval onto [itex]-\frac{\pi}{2},\frac{\pi}{2}[/itex].

    In this case, we have [tex]\frac{d}{dx}arcsine(x)=\frac{1}{\sqrt{1-x^{2}}}[/tex], and it follows that the anti-derivative will use the +sign
     
  6. Dec 29, 2009 #5

    dextercioby

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    You may wonder where the range of "arcsin" comes into play. Remember that

    [tex] \frac{d \mbox{arcsin} x}{dx} = \pm \frac{1}{\sqrt{1-x^2}} [/tex], because

    in the computation of this derivative you meet a point where

    [tex] \sin x = p \Rightarrow \cos x = \pm \sqrt{1-p^2} [/tex].

    To remove the sign ambiguity, you have to choose an appropriate interval either for the domain or for the range of the functions involved.
     
  7. Dec 29, 2009 #6
    Thanks for the replies arildno and bigubau.

    @Bigubau, so do you mean this step when differentiating arcsin(x):

    let [tex]y= arcsin(x) , sin(y)=x[/tex] and

    differentiating [tex]sin(y)= x[/tex],

    [tex]cos(y) \frac{dy}{dx} =1 [/tex]

    => [tex]\frac{dy}{dx}=\frac{1}{cos(y)}=*\frac{1}{\sqrt{1-sin^2(y)}}=*\frac{1}{\sqrt{1-x^2}}[/tex]

    * where it should be [tex]cos(y)= \pm\sqrt{1-sin^2(y)}=\pm\sqrt{1-x^2}[/tex]

    When I integrate arcsin(x) using integration by parts, I get

    [tex] xarcsin(x) - \int \frac{x}{\sqrt{1-x^2}} dx [/tex]

    so it comes out with the positive [tex]+\sqrt{1-x^2}[/tex]

    Is there another method to integrate the last part? Apparently, when I calculate a definite integral, it's got to be either plus or minus and my calculator and textbook demands the minus. For example they both give

    [tex]\int_{0}^{1} arcsin(x) = \frac{\pi}{2}-1[/tex]



    Thanks for the support,
    Charismaztex
     
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