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Math Help - Integration by parts

  1. #1
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    Integration by parts

    I went wrong somewhere here. First let's say:
    L=\int e^{-5x}sin(x)dx

    Let's integrate that by parts. Let's start with:
    u=e^{-5x}
    u'=-5e^{-5x}
    v=-cosx
    v'=sinx

    Next:
    (e^{-5x})(-cosx-\int(-5e^{-5x})(-cosx)dx
    (e^{-5x})(-cosx+5\int(e^{-5x})(-cosx)dx

    Let's integrate by parts again on that new integral because I think I can get back to the original integral:
    \int(e^{-x})(-cosx)dx
    u=-cosx
    u'=sinx
    v=-\frac{1}{5}e^{-5x}
    v'=e^{-5x}

    So we have:
    (-cosx)(-\frac{1}{5}e^{-5x})-\int(sinx)(-\frac{1}{5}e^{-5x})
    (-cosx)(-\frac{1}{5}e^{-5x})+\frac{1}{5}\int(e^{-5x})(sinx)

    All together:
    (e^{-5x})(-cosx)+5\left [(-cosx)(-\frac{1}{5}e^{-5x})+\frac{1}{5}L\right ]=L

    I don't think I did it right because the my next step appears to be a dead end:
    -e^{-5x}cosx+e^{-5x}+L=L

    I did something wrong with the coefficients or the signs methinks... or everything. That's always a possibility with me.
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Integration by parts

    Quote Originally Posted by jamesrb View Post
    I don't think I did it right because the my next step appears to be a dead end:
    -e^{-5x}cosx+e^{-5x}cosx+L=L
    0 = 0
    /edited.

    so while you were right in your process, this yielded no information. Let u=e^{-5x} and v' = -cosx instead for your second integral.
    Thanks from jamesrb
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  3. #3
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    Re: Integration by parts

    Quote Originally Posted by MacstersUndead View Post
    /edited.

    so while you were right in your process, this yielded no information. Let u=e^{-5x} and v' = -cosx instead for your second integral.
    Alright so if I do that I get:
    (e^{-5x})(-sinx)-\int(-5e^{-5x})(-sin(x)
    (e^{-5x})(-sinx)+5\int(e^{-5x})(-sin(x)

    Can I pull a -1 out of [tex](-sin(x))[/text]? To get:
    (e^{-5x})(-sinx)-5\int(e^{-5x})(sin(x)

    So all together:
    (e^{-5x})(-cosx)+5\left [(e^{-5x})(-sinx)-5L\right ]=L
    -e^{-5x}cosx-5e^{-5x}cosx-25L=L
    -e^{-5x}cosx-5e^{-5x}cosx=26L

    Finally:
    \frac{-e^{-5x}cosx-5e^{-5x}sinx}{26}

    My calculus book says the answer is:
    -\frac{1}{26}e^{-5x}(cosx+5sinx)+C

    I think those are equivalent answers.
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  4. #4
    Senior Member MacstersUndead's Avatar
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    Re: Integration by parts

    They are equivalent answers. Well done.
    Thanks from jamesrb
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