Answer
$$\frac{1}{2}[e^t-\cos t-\sin t]$$
Work Step by Step
Given
$$I=\int_{0}^{t}e^s\sin(t-s)ds$$
Let
\begin{align*}
u&=e^s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sin(t-s)ds\\
u&=e^sds\ \ \ \ \ \ \ \ \ \ \ \ \ v=\cos (t-s)
\end{align*}
Then using integration by parts
\begin{align*}
I&=uv-\int vdu\\
&= e^s\cos (t-s) \bigg|_{0}^{t}-\int_{0}^{t} e^s\cos (t-s)ds \\
&=e^t-\cos t-J
\end{align*}
where $J=\displaystyle\int_{0}^{t} e^s\cos (t-s)ds$
Let
\begin{align*}
u&=e^s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos (t-s) ds\\
u&=e^sds\ \ \ \ \ \ \ \ \ \ \ \ \ v=-\sin (t-s)
\end{align*}
then
\begin{align*}
J&= -e^s\sin (t-s) \bigg|_{0}^{t}+\int_{0}^{t} e^s\sin (t-s)ds\\
&=\sin t +I
\end{align*}
Hence
\begin{align*}
I &=e^t-\cos t-(\sin t+I)\\
&=\frac{1}{2}[e^t-\cos t-\sin t]
\end{align*}
