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.