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Integral of arcsin x

  1. Feb 22, 2004 #1
    I find it really hard to solve
    [tex]\int\arcsin{x}\;dx[/tex]
    I tried using integration by part with these values:
    [tex]
    \begin{array}{rl}
    u=\arcsin{x},&dv=dx\\
    du=\frac{1}{\sqrt{1-x^2}}\;dx,&v=x\\
    \end{array}
    [/tex]
    Which yields
    [tex]
    \begin{equationarray}
    \int\arcsin{x}\;dx&=&x\arcsin{x}-\int\frac{x}{\sqrt{1-x^2}}\;dx
    \end{equationarray}
    [/tex]
    After that I tried various substitutions/integration by parts but didn't get any simpler form... Can anyone help my desperate self :) (maybe hints...)

    Thanks a lot beforehand...
     
  2. jcsd
  3. Feb 22, 2004 #2

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    What substitutions did you try? There are at least two substitutions that work well here; one just screams "try me", and another is one of the techniques I think you've learned by now that is specifically for solving integrals that look like that...
     
    Last edited: Feb 22, 2004
  4. Feb 22, 2004 #3
    Try the substitution [itex] 1 - x^2 [/itex].

    Edit: Heh, guess I gave it away.

    cookiemonster
     
    Last edited: Feb 22, 2004
  5. Feb 22, 2004 #4
    Given: [tex] y=\arcsin x[/tex]

    Take the sin of both sides of the equation yielding: [tex]\sin y=x[/tex]

    Use implicit differentiation: [tex](\sin y)\frac{d}{dx}=(x)\frac{d}{dx}[/tex]

    to get: [tex](\cos y)dy=dx[/tex]

    Stick the above into this: [tex]\int\arcsin{x}\;dx[/tex]

    Substitute y, and (sin y)dy into your original equation: [tex]\int y\cos y\;dy[/tex]

    Integrate by parts: [tex]y\sin y+\cos y+C[/tex]

    Substitute x and arcsin x back into equation and replace cos y with the sqrt of 1- sin^2 y: [tex]x\arcsin x+\sqrt{1-sin^2y}+C[/tex]

    Substitute x for sin y: [tex]x\arcsin x+\sqrt{1-x^2}+C[/tex]

    Thus: [tex]\int\arcsin{x}\;dx=x\arcsin x+\sqrt{1-x^2}+C[/tex]

    You do pretty much the same as above for the other inverse trig functions. Hope this helped.

    You where going down the right path the way you chose to do it also. if you let u=1-x^2 then du would equal? Which could then be substituted back in to get the same answer as above...
     
  6. Feb 22, 2004 #5
    Ouch, I should've been able to see that (the u = 1-x^2 substitution) :). Btw your method is very interesting faust9.

    Thanks all :)
     
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