Answer
$$\frac{e^{\pi}+1}{2}$$
Work Step by Step
Given $$\int_{0}^{\pi} e^{x} \sin x d x$$
\begin{align*}
u &= \sin x \ \ \ \ \ \ \ \ \ dv= e^xdx\\
du &= \cos x dx \ \ \ \ \ \ \ \ v= e^x
\end{align*}
Then,
\begin{align*}
\int \sin x e^{x} d x &=(\sin x)\left(e^{x}\right)-\int e^{x} \cos x d x \\
&=e^{x} \sin x- \int \cos x e^{x} d x
\end{align*}
For the integral part again, let
\begin{align*}
u &= \cos x\ \ \ \ \ \ \ \ \ dv= e^xdx\\
du &= -\sin xdx \ \ \ \ \ \ \ \ v= e^x
\end{align*}Then,
\begin{align*}
\int \sin x e^{x} d x&=e^{x} \sin x- \left(e^{x} \cos x+ \int e^{x} \sin x d x\right)\\
\int \sin x e^{x} d x&=e^{x} \sin x- e^{x} \cos x- \int e^{x} \sin x d x\\
2 \int \sin x e^{x} d x&=e^{x} \sin x- e^{x} \cos x\\
\int \sin 2 x e^{x} d x&=\frac{1}{2} e^{x}(\sin x- \cos x) +C\\
\int _{0}^{\pi}\sin 2 x e^{x} d x&=\frac{1}{2} e^{x}(\sin x- \cos x) \bigg|_{0}^{\pi}\\
&= \frac{e^{\pi}+1}{2}
\end{align*}
