1. We are asked to evaluate the integral $$\int \sin^2 x \cos^3 x \, dx$$. 2. To solve this integral, we use trigonometric identities and substitution. Important rules: - Express powers of sine or cosine in terms of the other function if possible. - Use substitution when the derivative of one function appears in the integral. 3. Rewrite $$\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x$$ using the Pythagorean identity $$\cos^2 x = 1 - \sin^2 x$$. 4. Substitute into the integral: $$\int \sin^2 x \cos^3 x \, dx = \int \sin^2 x (1 - \sin^2 x) \cos x \, dx$$. 5. Let $$u = \sin x$$, then $$du = \cos x \, dx$$. 6. The integral becomes: $$\int u^2 (1 - u^2) \, du = \int (u^2 - u^4) \, du$$. 7. Integrate term-by-term: $$\int u^2 \, du - \int u^4 \, du = \frac{u^3}{3} - \frac{u^5}{5} + C$$. 8. Substitute back $$u = \sin x$$: $$\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$$. Final answer: $$\int \sin^2 x \cos^3 x \, dx = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$$.