1. **State the problem:** Evaluate the integral $$\int \cos^3(x) e^{\log(\sin x)} \, dx$$. 2. **Simplify the integrand:** Recall that $e^{\log(a)} = a$ for $a > 0$. So, $$e^{\log(\sin x)} = \sin x.$$ Therefore, the integral becomes $$\int \cos^3(x) \sin x \, dx.$$ 3. **Rewrite the integral:** $$\int \cos^3(x) \sin x \, dx = \int \cos^3(x) \sin x \, dx.$$ 4. **Use substitution:** Let $$u = \cos x,$$ then $$du = -\sin x \, dx,$$ or equivalently, $$-du = \sin x \, dx.$$ 5. **Substitute in the integral:** $$\int \cos^3(x) \sin x \, dx = \int u^3 (-du) = -\int u^3 \, du.$$ 6. **Integrate:** $$-\int u^3 \, du = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C.$$ 7. **Final answer:** $$\boxed{-\frac{\cos^4 x}{4} + C}.$$