1. **State the problem:** Evaluate the integral $$\int \sin(2x) e^{\cos(2x)} \, dx$$. 2. **Recall the formula and substitution rule:** When integrating a function of the form $$f(g(x)) g'(x)$$, substitution is useful. Here, notice that the derivative of $$\cos(2x)$$ is related to $$\sin(2x)$$. 3. **Set the substitution:** Let $$u = \cos(2x)$$. 4. **Compute the differential:** $$\frac{du}{dx} = -2 \sin(2x) \implies du = -2 \sin(2x) dx$$. 5. **Rewrite the integral:** From $$du = -2 \sin(2x) dx$$, we get $$\sin(2x) dx = -\frac{1}{2} du$$. 6. **Substitute into the integral:** $$\int \sin(2x) e^{\cos(2x)} dx = \int e^u \left(-\frac{1}{2} du\right) = -\frac{1}{2} \int e^u du$$. 7. **Integrate:** $$-\frac{1}{2} \int e^u du = -\frac{1}{2} e^u + C$$. 8. **Back-substitute:** $$-\frac{1}{2} e^{\cos(2x)} + C$$. **Final answer:** $$\boxed{-\frac{1}{2} e^{\cos(2x)} + C}$$