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Integration to arcsin.

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem, from the very beginning, was:

    [tex]\int \sqrt{9-x^{2}}dx[/tex]

    But this I have reduced to:

    [tex]\int \sqrt{9-x^{2}}dx = \frac{x}{2} \sqrt{9-x^{2}} + \frac{9}{2} \int \frac{1}{\sqrt{9-x^{2}}}dx [/tex]

    My problem is that last integral - I get a factor of (1/3) times the correct answer and I don't know what to do - I simply can't see it.
    2. Relevant equations

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

    3. The attempt at a solution

    I look at it and want to "transform" my expression into something like the arcsin expression above. So I say:

    [tex]\int \frac{1}{\sqrt{9-x^{2}}}dx = \int \frac{1}{3\sqrt{1-\frac{x^{2}}{9}}}dx[/tex]

    and from there get:

    [tex]\int \frac{1}{\sqrt{9-x^{2}}}dx = \frac{1}{3}arcsin\frac{x}{3}[/tex]

    but you're not supposed to get that factor of 1/3 - you're not supposed to remove it? Can anyone explain to me what I'm missing? I've searched my textbook so many times now I'm about to throw it into the wall or something..
  2. jcsd
  3. Oct 20, 2009 #2
    [tex]\int \frac{{\rm d}u}{\sqrt{1-u^2}}=\arcsin u[/tex]

    but here we have

    [tex]\int \frac{{\rm d}x}{\sqrt{1-\left(\frac x 3 \right)^2}}[/tex]

    [tex]u= \frac x 3 \Rightarrow {\rm d}x= ?[/tex]
  4. Oct 20, 2009 #3
    You need to use that relevant equation to deal with [tex]\frac{x}{3}[/tex] and not just [tex]x[/tex]. Be careful with that substitution!

    Also, a more natural way to deal with the original integral would be to consider the substitution [tex]x=3 \sin \theta [/tex] and go from there using some trig identities
  5. Oct 20, 2009 #4
    Ohh, I think I get it now, thank you! I'd say it needs dx/3, so my 1/3 "disappears" (for lack of better wording).

    Anyway, as for that last comment - you're probably right, but I don't think I've seen that type of thing being used before.
  6. Oct 20, 2009 #5
    No need to worry really. You will come across it sooner or later and it's good to know
  7. Oct 20, 2009 #6
    You could also have used the trig substitution x = 3sinθ which should make the integral pretty easy to work with.
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