We are asked to evaluate the following definite integral:

\[
\int_{-\pi}^{\pi} \sin(x) \cos(x) \, dx
\]

### Step 1: Simplify the integrand using a trigonometric identity

Recall the trigonometric identity:

\[
\sin(2x) = 2 \sin(x) \cos(x)
\]

Thus, we can rewrite the product of \(\sin(x)\) and \(\cos(x)\) as:

\[
\sin(x) \cos(x) = \frac{1}{2} \sin(2x)
\]

### Step 2: Substitute this into the integral

Now, the integral becomes:

\[
\int_{-\pi}^{\pi} \sin(x) \cos(x) \, dx = \frac{1}{2} \int_{-\pi}^{\pi} \sin(2x) \, dx
\]

### Step 3: Evaluate the integral

The integral of \(\sin(2x)\) is straightforward:

\[
\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)
\]

Now, apply the limits of integration:

\[
\frac{1}{2} \left[ -\frac{1}{2} \cos(2x) \right]_{-\pi}^{\pi}
\]

### Step 4: Evaluate the boundary values

At \(x = \pi\), we have:

\[
\cos(2\pi) = 1
\]

At \(x = -\pi\), we have:

\[
\cos(-2\pi) = 1
\]

So, the evaluation becomes:

\[
\frac{1}{2} \left[ -\frac{1}{2} (1 - 1) \right] = 0
\]

### Final Answer:

\[
\int_{-\pi}^{\pi} \sin(x) \cos(x) \, dx = 0
\]

Let me know if you want more details or have any questions!

Here are 5 related questions you might explore:
1. How would the integral change if the limits were from 0 to \(\pi\)?
2. Can you find the integral of \(\sin^2(x)\) over \([- \pi, \pi]\)?
3. What happens when integrating \(\cos^2(x)\) over \([- \pi, \pi]\)?
4. How do we approach integrals of products of sine and cosine functions in general?
5. Can you derive the identity for \(\sin(2x)\) using the sum of angles formula?

**Tip**: Always look for trigonometric identities when dealing with products of sine and cosine functions to simplify the integrand.

Explore how to evaluate the definite integral of sin(x)cos(x) over the limits from -π to π using trigonometric identities and integration techniques. This step-by-step guide covers key trigonometric concepts and is suitable for students in Grades 10-12.

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