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Thread: Integration by parts

  1. #1
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    Integration by parts

    Hi, I just learned integration by parts. I'm trying to solve:
     \int x^2 sin(3x+1)dx\
    I know usually you want u to have the simpler derivative.
    I set u=sin(3x+1), du=3cos(3x+1), v=(1/3)x^3, dv=x^2

    So using integration by parts, uv- integral vdu
    sin(3x+1)(1/3)x^3 - \int (1/3)x^3 3cos(3x+1)

    then I went through the steps to get my final answer:
    (1/3)x^3sin(3x+1)-(1/12)x^4sin(3x+1) +C

    Was my approach right? I wasn't too sure about this one...any help would be great.
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  2. #2
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    Quote Originally Posted by bcahmel View Post
    Hi, I just learned integration by parts. I'm trying to solve:
     \int x^2 sin(3x+1)dx\
    I know usually you want u to have the simpler derivative.
    I set u=sin(3x+1), du=3cos(3x+1), v=(1/3)x^3, dv=x^2

    So using integration by parts, uv- integral vdu
    sin(3x+1)(1/3)x^3 - \int (1/3)x^3 3cos(3x+1)

    then I went through the steps to get my final answer:
    (1/3)x^3sin(3x+1)-(1/12)x^4sin(3x+1) +C

    Was my approach right? I wasn't too sure about this one...any help would be great.
    try again ... note that you'll have to use parts more than once.

    u = x^2

    dv = \sin(3x+1) \, dx
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  3. #3
    Super Member TheChaz's Avatar
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    There is a rule for how to pick u and dv so that you won't be going in circles (as much, at least!).
    It's called the LIATE rule.
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  4. #4
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    thanks, for the LIATES rule and thanks to skeeter for reminding me that i use IBP more than once...
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