1. The problem is to evaluate the integral $$\int \frac{\sin(e^{-x})}{2e^x} \, dx$$. 2. We consider using substitution (u-substitution) to simplify the integral. 3. Let $$u = e^{-x}$$. Then, $$\frac{du}{dx} = -e^{-x} = -u$$, so $$du = -u \, dx$$ or $$dx = \frac{du}{-u}$$. 4. Rewrite the integral in terms of $$u$$: $$\int \frac{\sin(u)}{2e^x} dx$$. 5. Since $$u = e^{-x}$$, then $$e^x = \frac{1}{u}$$. 6. Substitute $$e^x = \frac{1}{u}$$ and $$dx = \frac{du}{-u}$$: $$\int \frac{\sin(u)}{2 \cdot \frac{1}{u}} \cdot \frac{du}{-u} = \int \frac{\sin(u)}{2/u} \cdot \frac{du}{-u} = \int \frac{\sin(u) \cdot u}{2} \cdot \frac{du}{-u}$$. 7. Cancel $$u$$ in numerator and denominator: $$\int \frac{\sin(u) \cancel{u}}{2} \cdot \frac{du}{-\cancel{u}} = \int \frac{\sin(u)}{2} \cdot (-1) du = -\frac{1}{2} \int \sin(u) du$$. 8. Integrate $$\sin(u)$$: $$\int \sin(u) du = -\cos(u) + C$$. 9. Substitute back: $$-\frac{1}{2} (-\cos(u)) + C = \frac{1}{2} \cos(u) + C = \frac{1}{2} \cos(e^{-x}) + C$$. Final answer: $$\boxed{\frac{1}{2} \cos(e^{-x}) + C}$$