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Another trig sub problem

  1. Jun 9, 2006 #1
    heres what i have:

    [tex] \int_{0}^{2\sqrt{3}} \frac {x^3}{\sqrt{16-x^2}}dx[/tex]
    let [tex]x=4sin\phi[/tex]


    [tex]\sqrt{16-x^2} = \sqrt{16-4sin^2\phi} = \sqrt{4cos^2\phi = 4cos\phi[/tex]

    [tex]\int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}dx[/tex]

    [tex]\int_{0}^{2\sqrt{3}} \frac{4sin^3\phi}{4cos\phi}4cos\phi d\phi[/tex]

    [tex]= 4\int_{0}^{2\sqrt{3}} sin^3/phi [/tex]


    [tex]=4\int_{0}^{2\sqrt{3}} 1-cos^2\phi(sin\phi)[/tex]

    [tex]=4\int_{0}^{2\sqrt{3}} sin\phi - sin \phi (cos^2\phi) [/tex]

    this is as far as i went because i think i messed up

    any help would be appreciated
    Last edited: Jun 9, 2006
  2. jcsd
  3. Jun 9, 2006 #2
    I don't see you changing your limits of integration when you made the substitution. To address your question though, back up one step and think to yourself "can I make a 2nd substitution that would greatly simplify the problem?".
  4. Jun 9, 2006 #3
    [tex]=4\int_{0}^{2\sqrt{3}} 1-cos^2\phi(sin\phi)[/tex]

  5. Jun 9, 2006 #4
    Disregarding the missing parentheses, I believe you could do something with that in the form of a substitution (well the same thing is visible in the last one but I felt it was slightly less obvious there)
  6. Jun 9, 2006 #5


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    A little help

    now let [tex]u=\cos\phi \Rightarrow du=-\sin\phi d\phi[/tex] and [tex]0\leq\phi\leq \frac{\pi}{3} \Rightarrow 1\leq u\leq \frac{1}{2}[/tex] to get

    [tex]=-64\int_{1}^{\frac{1}{2}} (1-u^2)du =64\int_{\frac{1}{2}}^{1} (1-u^2)du[/tex]
    [tex]= 64 \left[ u-\frac{1}{3}u^3\right]_{u = \frac{1}{2}}^{1} = 64 \left[ (1-\frac{1}{3}) - (\frac{1}{2}-\frac{1}{24} )\right] [/tex]
    [tex]=64 \left( \frac{2}{3} - \frac{11}{24}\right) = 64 \frac{5}{24} =\boxed{\frac{40}{3}}[/tex]

    The big help here was changing the bounds of integration with each substitution (rather thatn backwards substitution later).

    You're doing well: keep going.

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