= Solution
Let $I_n=\int e^{nx}\sin x\,dx$ and $J_n=\int e^{nx}\cos x\,dx$. Integrating the <trigonometric functions> and differentiating $e^{nx}$ in <integration by parts> gives, modulo constants,
$$
I_n=-e^{nx}\cos x+nJ_n,\qquad J_n=e^{nx}\sin x-nI_n.
$$
Substitution yields $(1+n^2)I_n=e^{nx}(n\sin x-\cos x)$ and $(1+n^2)J_n=e^{nx}(n\cos x+\sin x)$. Therefore
$$
\begin{aligned}
\int e^{nx}(\sin x-\cos x)\,dx&=I_n-J_n\\
&=\boxed{-\frac{e^{nx}}{n^2+1}\left[(1+n)\cos x+(1-n)\sin x\right]+C.}
\end{aligned}
$$
No division by $n$ was used, so this formula also applies at $n=0$.
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