Integrating by parts twice and coming back

First pass. With $u=\sin x$, $v'=e^x$:
$$J=\bigl[e^x\sin x\bigr]_0^\pi-\int_0^\pi e^x\cos x\,dx=-K,\qquad K=\int_0^\pi e^x\cos x\,dx.$$

Second pass. With $u=\cos x$, $v'=e^x$:
$$K=\bigl[e^x\cos x\bigr]_0^\pi+\int_0^\pi e^x\sin x\,dx=(-e^\pi-1)+J.$$

Solving. $J=-K=e^\pi+1-J$, so $2J=e^\pi+1$ and
$$J=\frac{e^\pi+1}2\approx12.07.$$

Remark. An antiderivative is $\frac{e^x(\sin x-\cos x)}2$. The "cyclic" method also works for $\int e^{ax}\cos bx\,dx$ and is the real-variable counterpart of taking the imaginary part of $\int e^{(1+i)x}dx$.