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Integral, Trig Substitution - Stuck

  1. Aug 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Use trigonometric substitution to evaluate


    3. The attempt at a solution

    Let [tex]x=3\sin\theta[/tex]
    then [tex]dx=3\cos\theta d\theta[/tex]


    [tex]=\int{\frac{9\sin^2\theta}{3\sqrt{1-\sin^2\theta}}}\ 3\cos\theta \ d\theta[/tex]

    [tex]=\int{\frac{9\sin^2\theta}{3\sqrt{\cos^2\theta}}}\ 3\cos\theta \ d\theta[/tex]

    [tex]=\int{9\sin^2\theta \ d\theta}[/tex]

    [tex]=\frac{9}{2} \int{(1-\cos2\theta)}\ d\theta[/tex]

    let [tex]w=2\theta[/tex]
    then [tex]dw=2\ d\theta[/tex]

    [tex]=\frac{9}{4} \int{(1-\cos w)}\ dw[/tex]

    [tex]=\frac{9}{4}[w-\sin w] + c \ \mbox{(substituting everything back in)}[/tex]


    Is this correct so far? And if so, now what?

    I'm stumped :redface:

  2. jcsd
  3. Aug 6, 2008 #2
    Your integration seems fine. I would clean up your substitution at the end. Use:

    [tex]sin2 \theta = 2 sin \theta cos \theta[/tex]
  4. Aug 6, 2008 #3
    Well at this step:

    [tex] =\frac{9}{2} \int{(1-\cos2\theta)}\ d\theta [/tex]

    you could just find the antiderivative rather easily without another substitution. Then you can just relate theta to x instead of relating w to theta to x.

    After you get the antiderivative in terms of x, that's pretty much it. Take the derivative to check.
  5. Aug 7, 2008 #4
    That's exactly what I needed. Thank you very much! :smile:
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