# Integration by parts

• Mar 2nd 2013, 06:42 PM
jamesrb
Integration by parts
I went wrong somewhere here. First let's say:
$\displaystyle L=\int e^{-5x}sin(x)dx$

$\displaystyle u=e^{-5x}$
$\displaystyle u'=-5e^{-5x}$
$\displaystyle v=-cosx$
$\displaystyle v'=sinx$

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.
• Mar 2nd 2013, 07:55 PM
Re: Integration by parts
Quote:

Originally Posted by jamesrb
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.
• Mar 2nd 2013, 08:24 PM
jamesrb
Re: Integration by parts
Quote:

/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.
• Mar 2nd 2013, 08:52 PM