1. **State the problem:** Find the definite integral $$\int_0^{\frac{\pi}{2}} \cos x \cdot e^{-\sin x} \, dx$$. 2. **Formula and substitution:** We notice the integrand is a product of $$\cos x$$ and an exponential function involving $$\sin x$$. A useful substitution is: $$u = -\sin x$$ Then, $$\frac{du}{dx} = -\cos x \implies du = -\cos x \, dx$$ So, $$\cos x \, dx = -du$$. 3. **Change the limits:** When $$x=0$$, $$u = -\sin 0 = 0$$. When $$x=\frac{\pi}{2}$$, $$u = -\sin \frac{\pi}{2} = -1$$. 4. **Rewrite the integral:** $$\int_0^{\frac{\pi}{2}} \cos x \cdot e^{-\sin x} \, dx = \int_{u=0}^{u=-1} e^u (-du) = \int_0^{-1} -e^u \, du = \int_{-1}^0 e^u \, du$$. 5. **Evaluate the integral:** $$\int_{-1}^0 e^u \, du = \left[ e^u \right]_{-1}^0 = e^0 - e^{-1} = 1 - \frac{1}{e}$$. 6. **Final answer:** $$\boxed{1 - \frac{1}{e}}$$. This is the exact value of the definite integral.