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The integral provided is:

\[
\int \sin^2(x) \cos^3(x) \, dx
\]

To solve this integral, we can use substitution and trigonometric identities. Here's a step-by-step solution:

### Step 1: Break down the powers of sine and cosine

We can express \(\cos^3(x)\) as:

\[
\cos^3(x) = \cos^2(x) \cdot \cos(x)
\]

Now recall the Pythagorean identity:

\[
\cos^2(x) = 1 - \sin^2(x)
\]

Thus, the integral becomes:

\[
\int \sin^2(x) (1 - \sin^2(x)) \cos(x) \, dx
\]

### Step 2: Substitution

Let \( u = \sin(x) \), so that \( du = \cos(x) \, dx \).

Substituting into the integral:

\[
\int u^2 (1 - u^2) \, du
\]

### Step 3: Expand the expression

Now expand the expression inside the integral:

\[
\int (u^2 - u^4) \, du
\]

### Step 4: Integrate

Now integrate each term:

\[
\int u^2 \, du = \frac{u^3}{3}, \quad \int u^4 \, du = \frac{u^5}{5}
\]

Thus, the integral becomes:

\[
\frac{u^3}{3} - \frac{u^5}{5} + C
\]

### Step 5: Substitute back \( u = \sin(x) \)

Substitute \( u = \sin(x) \) back into the result:

\[
\frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} + C
\]

### Final Answer:

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

Would you like more details or have any questions?  

Here are some related questions to deepen your understanding:
1. How do you handle integrals involving higher powers of sine and cosine?
2. Can you solve a similar integral with different powers, such as \(\int \sin^4(x) \cos^2(x) \, dx\)?
3. What is the advantage of using substitution in trigonometric integrals?
4. Could you approach this integral using double-angle or half-angle identities?
5. How would you integrate \(\int \cos^5(x) \, dx\)?

**Tip:** When working with integrals of trigonometric functions raised to powers, try to break down the powers using identities like the Pythagorean identity or by factoring one of the functions for easier substitution.

Learn how to solve the integral ∫ sin^2(x) cos^3(x) dx using substitution and trigonometric identities. This step-by-step solution covers key concepts in trigonometry and integration, ideal for Grades 11-12 students.

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