We are tasked with solving the integral:

\[
I = \int_0^\pi \cos^3(x) \sin(2x) \, dx
\]

### Step 1: Express \(\sin(2x)\) in terms of basic trigonometric functions
We use the identity \(\sin(2x) = 2\sin(x)\cos(x)\). So the integral becomes:

\[
I = \int_0^\pi \cos^3(x) \cdot 2\sin(x)\cos(x) \, dx
\]

This simplifies to:

\[
I = 2 \int_0^\pi \cos^4(x) \sin(x) \, dx
\]

### Step 2: Substitution
Let \( u = \cos(x) \), then \( du = -\sin(x)dx \). The limits change as follows:
- When \( x = 0 \), \( u = \cos(0) = 1 \)
- When \( x = \pi \), \( u = \cos(\pi) = -1 \)

Thus, the integral becomes:

\[
I = 2 \int_1^{-1} u^4 (-du)
\]

This simplifies to:

\[
I = 2 \int_{-1}^1 u^4 \, du
\]

### Step 3: Solving the integral
We now integrate \( u^4 \):

\[
I = 2 \left[ \frac{u^5}{5} \right]_{-1}^1
\]

Evaluating this:

\[
I = 2 \left( \frac{1^5}{5} - \frac{(-1)^5}{5} \right)
\]

\[
I = 2 \left( \frac{1}{5} - \frac{-1}{5} \right) = 2 \times \frac{2}{5} = \frac{4}{5}
\]

### Final Answer:
\[
\boxed{\frac{4}{5}}
\]

Would you like further clarification or details on any step?

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Here are some follow-up questions related to this topic:
1. How would you approach integrating \(\cos^n(x)\sin(2x)\) for a general \(n\)?
2. Can you find a different way to solve the integral using parts?
3. What is the geometric interpretation of integrating trigonometric functions over \([0, \pi]\)?
4. How would you solve an integral involving \(\cos^n(x)\sin(mx)\) for arbitrary \(n\) and \(m\)?
5. Can you apply a similar substitution to integrals involving \(\sin^n(x)\cos(2x)\)?

**Tip**: Trigonometric identities like \(\sin(2x)\) are extremely useful for simplifying integrals involving products of trigonometric functions.

This solution provides a step-by-step method for solving the integral ∫_0^π cos^3(x)sin(2x) dx using trigonometric identities and substitution. Learn how to simplify trigonometric expressions and integrate using basic calculus techniques. Suitable for undergraduate students.

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