integration by parts: integral( (e^-s*t)*sin(t) dt)

[rooney]

I'm trying to find a Laplace transform, and getting stuck in the integration. I have:

integral( (e^-s*t)*sin(t) dt)

i tried using u = e^-s*t and v' = sin(t) which gave me

(e^(-s*t))(-cos(t))-integral((-s*e^(-s*t))*-cos(t) dt)

i feel like i am back where i started.

 

[galactus]

Re: integration by parts

Hey Rooney:

$\displaystyle \int{e^{-st}sin(t)}dt$

Let 's leave $\displaystyle u=e^{-st}, \;\ dv=sin(t)dt, \;\ du=-se^{-st}dt, \;\ v=-cos(t)$

$\displaystyle -e^{-st}cos(t)-s\int{e^{-st}cos(t)}dt$

Now, leave $\displaystyle u=e^{-st}, \;\ dv=cos(t)dt, \;\ du=-se^{-st}dt, \;\ v=sin(t)$

$\displaystyle \int{e^{-st}sin(t)}dt=-e^{-st}cos(t)-s\left[e^{-st}sin(t)+s\int{e^{-st}sin(t)}dt\right]$

$\displaystyle \int{e^{-st}sin(t)}dt=-e^{-st}cos(t)-se^{-st}sin(t)-s^{2}\int{e^{-st}sin(t)}dt$

Now, see here?. It appears we're going it that circle you mentioned. But not really.

Add the integral on the right to the one on the left. See?. They are the same.

$\displaystyle \int{e^{-st}sin(t)}dt+s^{2}\int{e^{-st}sin(t)}dt=-e^{-st}cos(t)-se^{-st}sin(t)$

Factor on the left side:

$\displaystyle (1+s^{2})\int{e^{-st}sin(t)}dt=-e^{-st}cos(t)-se^{-st}sin(t)$

$\displaystyle \int{e^{-st}sin(t)}dt=\frac{-e^{-st}cos(t)-se^{-st}sin(t)}{s^{2}+1}$

You can tidy it up a bit if you wish. But if confronted with another integral like this, remember this technique.

It is common when you have 'cyclic' integrals like e and sin and cos.

 

[rooney]

Thanks, man. I appreciate it. I was tearing my hair out over this one.