1. **State the problem:** We want to solve the integral $$\int \sin^3(x) \, dx$$. 2. **Use the formula and rules:** Recall that $$\sin^3(x) = \sin(x) \cdot \sin^2(x)$$ and use the Pythagorean identity $$\sin^2(x) = 1 - \cos^2(x)$$. 3. **Rewrite the integral:** $$\int \sin^3(x) \, dx = \int \sin(x)(1 - \cos^2(x)) \, dx$$. 4. **Substitution:** Let $$u = \cos(x)$$, then $$du = -\sin(x) \, dx$$ or $$-du = \sin(x) \, dx$$. 5. **Rewrite integral in terms of $$u$$:** $$\int \sin(x)(1 - \cos^2(x)) \, dx = \int (1 - u^2)(-du) = -\int (1 - u^2) \, du$$. 6. **Integrate:** $$-\int (1 - u^2) \, du = -\left( u - \frac{u^3}{3} \right) + C = -u + \frac{u^3}{3} + C$$. 7. **Back-substitute $$u = \cos(x)$$:** $$-\cos(x) + \frac{\cos^3(x)}{3} + C$$. **Final answer:** $$\int \sin^3(x) \, dx = -\cos(x) + \frac{\cos^3(x)}{3} + C$$