1. We are asked to evaluate the integral $$\int \sin^2 x \cos^3 x \, dx$$. 2. To solve this, we use the substitution method and trigonometric identities. Notice that the powers of sine and cosine are different, so we try to express the integral in terms of one trigonometric function. 3. Rewrite $$\cos^3 x$$ as $$\cos x \cdot \cos^2 x$$ and use the identity $$\cos^2 x = 1 - \sin^2 x$$: $$\int \sin^2 x \cos^3 x \, dx = \int \sin^2 x \cos x (1 - \sin^2 x) \, dx$$. 4. Let $$u = \sin x$$, then $$du = \cos x \, dx$$. Substitute into the integral: $$\int u^2 (1 - u^2) \, du = \int (u^2 - u^4) \, du$$. 5. Integrate term-by-term: $$\int u^2 \, du - \int u^4 \, du = \frac{u^3}{3} - \frac{u^5}{5} + C$$. 6. Substitute back $$u = \sin x$$: $$\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$$. 7. Therefore, the integral evaluates to: $$\int \sin^2 x \cos^3 x \, dx = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C$$. This matches the top-right answer option and is the most accurate solution.