Solution
= Solution
Let $I=\int e^x\sin x\,dx$. Two applications of <integration by parts> give
$$
I=e^x\sin x-\int e^x\cos x\,dx,\qquad
\int e^x\cos x\,dx=e^x\cos x+I,
$$
where additive constants can be absorbed at the end. Solving for $I$ yields
$$
\boxed{\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x)+C.}
$$
Differentiating the answer gives $e^x\sin x$.