1. We are asked to evaluate the integral $$\int \sin x e^{\cos x} \, dx.$$\n\n2. Notice that the integrand contains $\sin x$ and $e^{\cos x}$. We can try substitution. Let $$u = \cos x.$$\n\n3. Then, $$\frac{du}{dx} = -\sin x \implies du = -\sin x \, dx.$$\n\n4. Rearranging, $$\sin x \, dx = -du.$$\n\n5. Substitute into the integral: $$\int \sin x e^{\cos x} \, dx = \int e^u (-du) = -\int e^u \, du.$$\n\n6. The integral of $e^u$ with respect to $u$ is $e^u$, so we get: $$-e^u + C.$$\n\n7. Substitute back $u = \cos x$: $$-e^{\cos x} + C.$$\n\n8. Therefore, the integral evaluates to $$-e^{\cos x} + C,$$ which corresponds to option (b).