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.
