1. The problem is to integrate the function $$\sin^3 x \cos^4 x$$ with respect to $$x$$. 2. Begin by rewriting $$\sin^3 x$$ as $$\sin x \cdot \sin^2 x$$. 3. Use the identity $$\sin^2 x = 1 - \cos^2 x$$ to express this as $$\sin x (1 - \cos^2 x) \cos^4 x$$. 4. So the integral becomes $$\int \sin x (1 - \cos^2 x) \cos^4 x \ dx$$. 5. Make the substitution $$u = \cos x$$, so $$du = -\sin x \ dx$$, or equivalently $$-du = \sin x \ dx$$. 6. Substitute into the integral: $$\int \sin x (1 - \cos^2 x) \cos^4 x \ dx = \int (1 - u^2) u^4 (-du) = - \int (1 - u^2) u^4 du$$. 7. Distribute inside the integral: $$- \int (u^4 - u^6) du = - \left( \int u^4 du - \int u^6 du \right)$$. 8. Integrate each term: $$- \left( \frac{u^5}{5} - \frac{u^7}{7} \right) + C = - \frac{u^5}{5} + \frac{u^7}{7} + C$$. 9. Substitute back $$u = \cos x$$: $$ - \frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C$$. Final answer: $$\int \sin^3 x \cos^4 x \ dx = - \frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C$$.