We are tasked with evaluating the following definite integral:

$I = \int_0^\pi \cos^2(x) \sin(2x) \, dx$

Step 1: Simplify the integrand

We can use a trigonometric identity to express $\sin(2x)$:

$\sin(2x) = 2\sin(x)\cos(x)$

Thus, the integrand becomes:

$\cos^2(x) \sin(2x) = \cos^2(x) \cdot 2\sin(x)\cos(x) = 2\cos^3(x) \sin(x)$

So the integral now becomes:

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

Step 2: Substitution

Let’s perform a substitution to make the integral easier. Let:

$u = \cos(x)$

Then:

$du = -\sin(x) \, dx$

The limits of integration will change as follows:

• When $x = 0$, $u = \cos(0) = 1$
• When $x = \pi$, $u = \cos(\pi) = -1$

Thus, the integral becomes:

$I = 2 \int_1^{-1} u^3 (-du) = 2 \int_{-1}^1 u^3 \, du$

Step 3: Evaluate the integral

We now evaluate the integral of $u^3$:

$\int u^3 \, du = \frac{u^4}{4}$

So, we now calculate:

$I = 2 \left[ \frac{u^4}{4} \right]_{-1}^1$

Substituting the limits:

$I = 2 \left( \frac{1^4}{4} - \frac{(-1)^4}{4} \right) = 2 \left( \frac{1}{4} - \frac{1}{4} \right)$

$I = 2 \times 0 = 0$

Final Answer:

$I = 0$

The value of the integral is $0$.

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Do you need more details or have any questions? Here are 5 related questions to consider:

1. How do trigonometric identities help simplify integrals?
2. What are other common substitution techniques for integrals?
3. How would this integral change if the limits were different?
4. Can you explain why the integral from $-1$ to 1 for an odd function is always zero?
5. How would you integrate $\cos^3(x) \sin(x)$ without using substitution?

Tip: Always look for symmetry or properties of odd and even functions to simplify definite integrals!