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Math Help - Integration by parts

  1. #1
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    Integration by parts

    \displaystyle\int(w^2-6w)sin(2w)dw

    u= (w^2-6w) du= (2w-6) dw

    dv= sin(2w)dw and v= \frac{-1}{2}cos(2w)

    \displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw

    u= (2w-6) du= 2 dw

    dv= cos(2w)dw and v= \frac{1}{2}sin(2w)

    \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\frac{1}{2}sin(2w)*2dw

    \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\sin(2w)dw

    \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{2}cos(2w)+c

    Is my algebra wrong? Cause the book is saying that the answer is:

    \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{4}cos(2w)+c
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    \displaystyle\int(w^2-6w)sin(2w)dw

    u= (w^2-6w) du= (2w-6) dw

    dv= sin(2w)dw and v= \frac{-1}{2}cos(2w)

    \displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw
    v=\frac{-1}{2}cos(2w) \;and\; u= (w^2-6w) \implies uv\;=\;\frac{-1}{2}cos(2w)\times  (w^2-6w)

    \displaystyle \int u\;dv = uv - \intv\;du = \dfrac{-(w^2-6w)\cos(2w)}{2} + \int \frac{1}{2}cos(2w)(2w-6) dw
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  3. #3
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    I'm sorry, I don't understand what you just did there. I thought the formula for the integration by parts is \displaystyle\int u\ dv=uv-\int v\ du
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    I'm sorry, I don't understand what you just did there. I thought the formula for the integration by parts is \displaystyle\int u\ dv=uv-\int v\ du
    It is..

    So look at your integration by parts.. You have written them down on the first and second lines of your answer..

    So what is uv??

    instead of writing uv= \displaystyle\frac{-1}{2}(w^2-6w)cos(2w), you have written uv=\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)
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  5. #5
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    Quote Originally Posted by dbakeg00 View Post
    \displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw

    u= (2w-6) du= 2 dw

    dv= cos(2w)dw and v= \frac{1}{2}sin(2w)

    \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\frac{1}{2}sin(2w)*2dw
    You are not handling the -\frac{1}{2} term properly in the integral in the top line. Since you are not including this when integrating by parts, I assume you are treating it like this:

    \displaystyle-\frac{1}{2}(w^2-6w)\sin(2w)+\frac{1}{2} \left[ \int (2w-6)\cos(2w) \, dw \right]

    After integrating by parts, you are making a substitution in place of the integral in the above line (the part in brackets). So, after your substitution, you get this:

    \displaystyle-\frac{1}{2}(w^2-6w)\sin(2w)+\frac{1}{2} \left[ \frac{1}{2}(2w-6)\sin(2w) - \int \sin(2w) \, dw \right]

    Does this help?
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  6. #6
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    Quote Originally Posted by harish21 View Post
    It is..

    So look at your integration by parts.. You have written them down on the first and second lines of your answer..

    So what is uv??

    instead of writing uv= \displaystyle\frac{-1}{2}(w^2-6w)cos(2w), you have written uv=\displaystyle\frac{-1}{2}(w^2-6w)sin(2w)
    What you are identifying was just a typo. Notice that he didn't carry that error through the remainder of his work.
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  7. #7
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    Ahh..I see it now. Thanks for helping me out guys...I really appreciate it!
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