1. **State the problem:** Evaluate the integral $$\int_0^\infty \frac{\sin^2(x) \cos(x)}{e^x} \, dx.$$\n\n2. **Rewrite the integral:** Note that $$e^x = \exp(x)$$ and the integral is improper with upper limit infinity. We want to find $I = \int_0^\infty \sin^2(x) \cos(x) e^{-x} \, dx$.\n\n3. **Use trigonometric identities:** Recall that $$\sin^2(x) = \frac{1 - \cos(2x)}{2}.$$ Substitute this into the integral:\n$$I = \int_0^\infty \frac{1 - \cos(2x)}{2} \cos(x) e^{-x} \, dx = \frac{1}{2} \int_0^\infty (1 - \cos(2x)) \cos(x) e^{-x} \, dx.$$\n\n4. **Distribute the integral:**\n$$I = \frac{1}{2} \left( \int_0^\infty \cos(x) e^{-x} \, dx - \int_0^\infty \cos(2x) \cos(x) e^{-x} \, dx \right).$$\n\n5. **Evaluate the first integral:**\nUse the formula for Laplace transform of cosine: $$\int_0^\infty e^{-sx} \cos(ax) \, dx = \frac{s}{s^2 + a^2}$$ for $s > 0$. Here, $s=1$, $a=1$, so\n$$\int_0^\infty \cos(x) e^{-x} \, dx = \frac{1}{1^2 + 1^2} = \frac{1}{2}.$$\n\n6. **Simplify the second integral:**\nUse product-to-sum identity: $$\cos(2x) \cos(x) = \frac{\cos(3x) + \cos(x)}{2}.$$\nSo,\n$$\int_0^\infty \cos(2x) \cos(x) e^{-x} \, dx = \frac{1}{2} \int_0^\infty (\cos(3x) + \cos(x)) e^{-x} \, dx.$$\n\n7. **Split the integral:**\n$$= \frac{1}{2} \left( \int_0^\infty \cos(3x) e^{-x} \, dx + \int_0^\infty \cos(x) e^{-x} \, dx \right).$$\n\n8. **Evaluate each integral:**\n- For $\cos(3x)$, $a=3$, $s=1$:\n$$\int_0^\infty \cos(3x) e^{-x} \, dx = \frac{1}{1^2 + 3^2} = \frac{1}{10}.$$\n- We already have $\int_0^\infty \cos(x) e^{-x} \, dx = \frac{1}{2}.$$\n\n9. **Sum the results:**\n$$\int_0^\infty \cos(2x) \cos(x) e^{-x} \, dx = \frac{1}{2} \left( \frac{1}{10} + \frac{1}{2} \right) = \frac{1}{2} \times \frac{6}{10} = \frac{3}{10}.$$\n\n10. **Substitute back into \(I$:**\n$$I = \frac{1}{2} \left( \frac{1}{2} - \frac{3}{10} \right) = \frac{1}{2} \times \frac{5}{10} = \frac{1}{4}.$$\n\n**Final answer:** $$\boxed{\frac{1}{4}}.$$