To evaluate the integral of $2 \sin^3(x) \cos^2(x) \, dx$, we can use a trigonometric identity and substitution. Let's go through it step-by-step.

1. Rewrite the integrand:
   Since $\sin^3(x) = \sin(x) \cdot \sin^2(x)$ and $\sin^2(x) = 1 - \cos^2(x)$, we can rewrite the integral as: $\int 2 \sin^3(x) \cos^2(x) \, dx = \int 2 \sin(x) (1 - \cos^2(x)) \cos^2(x) \, dx.$

2. Expand and simplify: $\int 2 \sin(x) \cos^2(x) - 2 \sin(x) \cos^4(x) \, dx.$

3. Substitute $u = \cos(x)$: Let $u = \cos(x)$, hence $du = -\sin(x) \, dx$ or $-du = \sin(x) \, dx$.

   The integral becomes: $\int -2u^2 \, du + \int 2u^4 \, du.$

4. Integrate: Integrating each term separately, $\int -2u^2 \, du = -2 \cdot \frac{u^3}{3} = -\frac{2u^3}{3},$ $\int 2u^4 \, du = 2 \cdot \frac{u^5}{5} = \frac{2u^5}{5}.$

5. Combine the results: $-\frac{2u^3}{3} + \frac{2u^5}{5} + C.$

6. Substitute back $u = \cos(x)$: $-\frac{2 \cos^3(x)}{3} + \frac{2 \cos^5(x)}{5} + C.$

So, the evaluated integral is: $\int 2 \sin^3(x) \cos^2(x) \, dx = -\frac{2 \cos^3(x)}{3} + \frac{2 \cos^5(x)}{5} + C.$

Would you like to go into more details or have any questions?

Here are some related questions that might help expand your understanding:

1. How can we evaluate an integral using trigonometric identities?
2. What other methods can we use for integrating powers of sine and cosine functions?
3. How does substitution simplify the integration process?
4. What are the common trigonometric substitutions used in integrals?
5. Can you derive the antiderivative for $\sin^n(x) \cos^m(x)$ for different values of $n$ and $m$?

Tip: When integrating trigonometric functions, look for identities or substitutions that can simplify the integrand.