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Antiderivative help

  1. Sep 28, 2004 #1
    can anyone help me find an antiderivative for (x^2)((9-(x^2))^(1/2))
     
  2. jcsd
  3. Sep 28, 2004 #2

    HallsofIvy

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    When you see [itex]\sqrt{1- x^2}[/itex] or anything like that, you should think [itex]cos(x)=\sqrt{1- sin^2(x)}[/itex]- and use a trig substitution.

    In this problem, factor a "9" out of the squareroot to get [itex]3x^2\sqrt{1- \frac{x^2}{9}}[/itex]. Now make the substitution x= 3sin(θ).

    dx= 3cos(θ)dθ and [itex]\sqrt{1- \frac{x^2}{9}}[/itex] becomes [itex]\sqrt{1- sin^2(\theta)}= cos(\theta)[/itex]. The entire integrand becomes sin2(θ)cos2(θ)dθ. You will need to use trig substitutions to integrate that.
     
  4. Sep 28, 2004 #3
    I think you may also be able to do it by parts if you let u=x and dv=the rest.
     
  5. Sep 29, 2004 #4
    Thanks alot, i was really just wondering if it was possible to do with the scope of my year 12 specialist maths course, and it seems as i can't. As that substitution i am not familiar with and intergration by parts is not on the course, is there another way with only linear substitution, substitution by "u" or partial fractions?
     
  6. Sep 29, 2004 #5

    Zurtex

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    There are a few different ways, you could use the + (a - a) method, by-parts and substitution. But at some point in all of them you are either going to have to make a trigonometric substitution or put it into standard form (which is also basically using a trig substitution but without the effort of workings). Quite simply because the anti derivative has inverse sine in it.
     
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