integrate integral from 0 to t of e^s sin(t-s)ds.
Hello twilightstrThis is one of these integration by parts questions where you integrate by parts twice, and this brings you back to the integral you started with (plus some other bits as well). If you call the original integral $\displaystyle I$, you can then put $\displaystyle I$ back into the equation and then solve the resulting equation for $\displaystyle I$.
Here's the first bit:
$\displaystyle I = \int_0^t e^s \sin(t-s)ds = [e^s \cos(t-s)]_0^t - \int_0^te^s \cos(t-s)ds$
Is that OK? I've integrated $\displaystyle \sin(t-s)$ to get $\displaystyle \cos(t-s)$, and differentiated $\displaystyle e^s$ to get $\displaystyle e^s$.
Now use Parts again on the second integral, and you'll find you get back to the integral you started with. Put this equal to $\displaystyle I$, and solve.
See if you can complete it now.
Grandad