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Math Help - [SOLVED] Integration by parts

  1. #1
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    [SOLVED] Integration by parts

    This is a problem I am having trouble with:
      <br />
\int 5xcos(5x) dx<br />

    This is the formula we were taught in class to use for these types of problems:
    <br />
\int udv = uv - \int vdu<br />

    I came up with this but I am sure it's wrong.
    <br />
xsin(5x) - \int sin(5x)<br />
    Last edited by redman223; August 31st 2008 at 02:35 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    This is a problem I am having trouble with:
      <br />
\int 5xcos(5x) dx<br />

    This is the formula we were taught in class to use for these types of problems:
    <br />
\int udv = uv - \int vdu<br />

    I came up with this but I am sure it's wrong.
    <br />
sin(5x) - \int xsin(5x)<br />
    yes, it's incorrect. did you identify your u and dv? show your steps so we see where you went wrong.
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  3. #3
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    I set u=5x du=5 dv=cos(5x) and v=1/5 sin(5x)

    That gave me:
    <br />
xsin(5x) - \int sin(5x)<br />

    I tried to use -1/5x cos(5x) for the part in the integral but it is still wrong.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    I set u=5x du=5 dv=cos(5x) and v=1/5 sin(5x)

    That gave me:
    <br />
xsin(5x) - \int sin(5x)~{\color{red}dx}<br />

    I tried to use -1/5x cos(5x) for the part in the integral but it is still wrong.
    that is correct. all you need to do now is find \int \sin 5x~dx and wrap it up
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  5. #5
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    Thanks, for some reason I kept thinking that the derivative of 5x was 5x and not 5, lol.

    I came up with -1/5 cos(5x) and it was correct. Thanks.
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