Answer
$\displaystyle \frac{e^{t}-\sin t-\cos t}{2}$
Work Step by Step
Integration by parts:
$\displaystyle \int udv=uv-\int vdu$
The idea is to choose a relatively easy $dv$ to integrate, and
a u whose $u'$ does not complicate matters (best case: makes thing simpler).
----
$\left[\begin{array}{ll}
u=\sin(t-s) & dv=e^{s}ds\\
& \\
du=-\cos(t-s)ds, & v=e^{s}
\end{array}\right]$
$I=\displaystyle \int_{0}^{t}e^{s}\sin(t-s)ds=uv|_{0}^{t}-\int_{0}^{t}vdu$
$= =[e^{s}\displaystyle \sin(t-s)]_{0}^{t}+\int_{0}^{t}e^{s}\cos(t-s)ds$
$=e^{t}\displaystyle \sin 0-e^{0}\sin t+\int_{0}^{t}e^{s}\cos(t-s)ds$
$=-\displaystyle \sin t+\int_{0}^{t}e^{s}\cos(t-s)ds$
By parts, again, $\left[\begin{array}{ll}
u=\cos(t-s) & dv=e^{s}ds\\
& \\
du=\sin(t-s)ds, & v=e^{s}
\end{array}\right]$
$\displaystyle \int_{0}^{t}e^{s}\cos(t-s)ds =[e^{s}\displaystyle \cos(t-s)]_{0}^{t}-\int_{0}^{t}e^{s}\sin(t-s)ds$
$=e^{t}\cos 0-e^{0}\cos t-I$
($I$ is the initial integral), so
$I=-\sin t+e^{t}-\cos t-I$
$2I=e^{t}-\sin t-\cos t$
$I=\displaystyle \frac{e^{t}-\sin t-\cos t}{2}$
