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Homework Help: Another Part where my brain's Fried

  1. May 7, 2005 #1
    [tex] \int \sqrt{4-x^2}^3 dx [/tex]

    I'm thinking integrating by parts would work, with u being that root and dv being dx, but is that the right method and direction. I've tried it and it seems more complicated.
  2. jcsd
  3. May 7, 2005 #2


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    Use [tex]x = 2 \sin \theta [/tex]...
    [tex]dx = 2 \cos \theta d\theta [/tex],
    and your integral becomes:
    [tex]16 \int \cos^4 \theta d\theta [/tex] .
    Use [tex]\cos{2\theta} = 2\cos^2 \theta - 1 [/tex], etc...
  4. May 7, 2005 #3
    Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!
  5. May 7, 2005 #4


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    So it's

    [tex]\int \left(4-x^{2}\right)^{\frac{3}{2}} \ dx[/tex]

    How about a substitution

    [tex] x=2\sin t [/tex] ?


    EDIT:Didn't see the other posts.
  6. May 7, 2005 #5


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    x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and
    [tex]\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)[/tex] so the integral becomes
    [tex]-16\int sin^4(u) du[/tex] and the only difference is that "-".
  7. May 7, 2005 #6
    Oops, I made a slight mistake!
  8. May 8, 2005 #7
    Thanks once more! :)
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