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Thread: Integration by parts

  1. #1
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    Integration by parts

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

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

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

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

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

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

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

    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
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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:
    $\displaystyle -e^{-5x}cosx+e^{-5x}cosx+L=L$
    $\displaystyle 0 = 0$
    /edited.

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

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

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

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

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

    I think those are equivalent answers.
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  4. #4
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    Re: Integration by parts

    They are equivalent answers. Well done.
    Thanks from jamesrb
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