To find the indefinite integral of $\sin^3(x) \cos^2(x) \, dx$, we can use trigonometric identities and substitution methods. Here’s a step-by-step solution:

Step 1: Rewrite the integrand using trigonometric identities

We know the identity: $\sin^2(x) = 1 - \cos^2(x)$ So we can rewrite $\sin^3(x)$ as: $\sin^3(x) = \sin(x) \cdot \sin^2(x) = \sin(x) \cdot (1 - \cos^2(x))$ Thus, the integral becomes: $\int \sin^3(x) \cos^2(x) \, dx = \int \sin(x) (1 - \cos^2(x)) \cos^2(x) \, dx$ Expanding the integrand: $\int \sin(x) \cos^2(x) \, dx - \int \sin(x) \cos^4(x) \, dx$

Step 2: Use substitution

Let $u = \cos(x)$, then $du = -\sin(x) \, dx$. The integrals become: $-\int u^2 \, du + \int u^4 \, du$ The negative sign comes from the substitution $du = -\sin(x) \, dx$.

Step 3: Integrate

Now integrate each term: $-\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cos^3(x)}{3}$ $\int u^4 \, du = \frac{u^5}{5} = \frac{\cos^5(x)}{5}$

Step 4: Combine the results

Combine the results to get the final answer: $\int \sin^3(x) \cos^2(x) \, dx = -\frac{\cos^3(x)}{3} + \frac{\cos^5(x)}{5} + C$ where $C$ is the constant of integration.

Final Answer

$\boxed{\int \sin^3(x) \cos^2(x) \, dx = -\frac{\cos^3(x)}{3} + \frac{\cos^5(x)}{5} + C}$

Would you like a detailed explanation of any particular step or have any questions?

Here are some related questions you might find interesting:

1. What is the indefinite integral of $\sin^2(x) \cos^3(x) \, dx$?
2. How would you integrate $\sin^4(x) \cos^2(x) \, dx$?
3. Can you find the indefinite integral of $\sin^2(x) \cos^4(x) \, dx$?
4. What substitution would you use for $\sin^3(x) \cos(x) \, dx$?
5. How would you solve $\int \sin(x) \cos(x) \, dx$ using a double-angle identity?

Tip: When integrating trigonometric functions raised to powers, always consider using identities or substitutions to simplify the expression.