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Integrate sqrt(9-x^2) by hand

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data

    $$\int_ \ \sqrt[2]{9-x^{2}} \, \mathrm{d} x.$$

    2. Relevant equations

    U substitution
    Integration by parts

    3. The attempt at a solution

    My teacher said it wasn't possible "by hand", but Wolframalpha provided a step-by-step solution. I'm clueless as to how wolfram came up with the substitutions/steps and how a mere mortal as myself could possibly be expected to do the same. If you could walk me through the logic behind the steps that wolfram provides, I would greatly appreciate it. I'm really just curious about the process as I just can't seem to wrap my head around it.

    Wolfram's steps: http://www.wolframalpha.com/widgets/view.jsp?id=dc816cd78d306d7bda61f6facf5f17f7

    Last edited: Jan 17, 2013
  2. jcsd
  3. Jan 17, 2013 #2


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    What do you mean 'by hand'. You can certainly integrate it with a trig substitution. Try x=3*sin(t). What's dx? The Wolfram reference you gave is about integrating sec^3(x). I don't see what that has to do with this problem.
    Last edited by a moderator: Jan 17, 2013
  4. Jan 17, 2013 #3
    Hey Dick,

    Thanks for your reply. I suppose we haven't learned trig substitution yet, so I guess that's my problem.
    As for the dx, doesn't that mean 'in respect to x'?

  5. Jan 17, 2013 #4


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    Not knowing trig substitution probably is not a help. But you know ordinary u substitution, yes? It's the same thing except you use trig functions. If you put x=t^2 then dx=2tdt. If you put x=3*sin(t) then dx=3*cos(t)dt. Same general idea.
  6. Jan 17, 2013 #5
    Alright then, I guess that's something I'll have to study later on.
  7. Jan 17, 2013 #6


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    Well, ok. You could probably do it now if you took a crack at it. It's not that mysterious. But if you want to wait till later, that's ok too.
  8. Jan 17, 2013 #7


    Staff: Mentor

    If it's a definite integral, with certain limits, then you can evaluate the integral be exploiting the geometry involved.

    For example, if it happens to be
    $$ \int_0^3 \sqrt{9 - x^2}dx$$
    then the integral represents one quarter of the area of a circle of radius 3, centered at (0, 0).
  9. Jan 18, 2013 #8


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    Integrate by parts

    $$\int \sqrt{a^2-x^2} \mathrm{dx}=
    x\sqrt{a^2-x^2}-\int \frac{-x^2}{\sqrt{a^2-x^2}} \mathrm{dx}$$
    $$=x \sqrt{a^2-x^2}-\int \frac{a^2-x^2}{\sqrt{a^2-x^2}} \mathrm{dx}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \mathrm{dx}=x \sqrt{a^2-x^2}-\int \sqrt{a^2-x^2} \mathrm{dx}+a^2\arcsin(x/a) $$
    From which we conclude
    $$\int \sqrt{a^2-x^2} \mathrm{dx}=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\arcsin(x/a)\right)$$
    An unrestricted form is
    $$\int \sqrt{a^2-x^2} \mathrm{dx}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+\arctan \left(\frac{x}{\sqrt{a^2-x^2}}\right)\right)
    Last edited by a moderator: Jan 18, 2013
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