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Math Help - Integral

  1. #1
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    Integral

    Hey,


    I am looking to antiderive the following:

    \int (3-sin{3x})(cos{3x})dx


    Any help is much appreciated,

    Dranalion
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  2. #2
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    \int (3-sin(3x))cos(3x)dx

    Just make the sub u=sin(3x), \;\ \frac{du}{3}=cos(3x)dx

    Now, it should be easier.

    \frac{1}{3}\int (3-u)du
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  3. #3
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    Using this, does this give the answer:

    = sin(3x) - 1/6(sin^2(3x)) + c ?

    I am just checking to see if I followed it correctly.
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  4. #4
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    Quote Originally Posted by Dranalion View Post
    Using this, does this give the answer:

    = sin(3x) - 1/6(sin^2(3x)) + c ?

    I am just checking to see if I followed it correctly.
    When you differentiate your answer do you get the integrand?
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  5. #5
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    Quote Originally Posted by Dranalion View Post
    Using this, does this give the answer:

    = sin(3x) - 1/6(sin^2(3x)) + c ?

    I am just checking to see if I followed it correctly.
    \frac{1}{3} \int(3-u)du = \frac{1}{3} (3u - \frac{u^2}{2}) + C = u - \frac{u^2}{6} + C = sin(3x) - \frac{sin^2(3x)}{6} + C so yes
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