1. ## Integration by parts

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

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

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.

2. ## Re: Integration by parts

Originally Posted by jamesrb
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.

3. ## Re: Integration by parts

/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.

4. ## Re: Integration by parts

They are equivalent answers. Well done.