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Math Help - [SOLVED] Trigonometric integration by parts #2

  1. #1
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    [SOLVED] Trigonometric integration by parts #2

    Using x=3sin\theta

    \int x^3 \sqrt {9-x^2} ~dx

    I think I solved the part under the radical correctly ( 3cos\theta)

    If I plug in x=3sin\theta and dx=3cos\theta and use 3cos\theta for the radical I get:

    \int 243sin^3\theta cos^2\theta ~d\theta

    I can change it to 243\int sin\theta(1-cos\theta)(cos^2\theta)

    But don't know what to do next.
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  2. #2
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    Quote Originally Posted by redman223 View Post
    Using x=3sin\theta

    \int x^3 \sqrt {9-x^2} ~dx

    I think I solved the part under the radical correctly ( 3cos\theta)

    If I plug in x=3sin\theta and dx=3cos\theta and use 3cos\theta for the radical I get:

    \int 243sin^3\theta cos^2\theta ~d\theta

    I can change it to 243\int sin\theta(1-cos\theta)(cos^2\theta)

    But don't know what to do next.
    Make the substitution u = \cos \theta.


    But you'd be much better off making the substitution u = 9 - x^2 in the original integral.

    You get - \frac{1}{2} \int (9 - u) \, \sqrt{u} \, du \, ......
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  3. #3
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    This is one of those problems where they want you to solve it using their method, so I am stuck using x=3sin\theta

    Which when I set u=cos\theta I come up with:

    -243(\frac {cos^3\theta}{3} - \frac {cos^4\theta}{4})+c

    If I am right, then what is the next step?
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  4. #4
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    Quote Originally Posted by redman223 View Post
    This is one of those problems where they want you to solve it using their method, so I am stuck using x=3sin\theta

    Which when I set u=cos\theta I come up with:

    -243(\frac {cos^3\theta}{3} - \frac {cos^4\theta}{4})+c

    If I am right, then what is the next step?
    Now you have to re-substitute x = 3 \sin \theta:

    From the Pythagorean Identity, \sin \theta = \frac{x}{3} \Rightarrow \cos \theta = \frac{\sqrt{9 - x^2}}{3} \, .....

    where it's sufficient to take the positive root.
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