The image shows a definite integral:

$\int_{-\pi}^{\pi} \sin(x) \, dx$

To solve this integral, we'll evaluate it step by step.

Step 1: Find the Antiderivative

The antiderivative of $\sin(x)$ is $-\cos(x)$.

$\int \sin(x) \, dx = -\cos(x)$

Step 2: Apply the Limits of Integration

Now, we apply the limits $-\pi$ and $\pi$ to the antiderivative.

$\left[-\cos(x)\right]_{-\pi}^{\pi} = -\cos(\pi) - \left(-\cos(-\pi)\right)$

Step 3: Evaluate the Cosine Function

We know that $\cos(\pi) = -1$ and $\cos(-\pi) = -1$. Substituting these values in:

$-\cos(\pi) - (-\cos(-\pi)) = -(-1) - (-(-1)) = 1 - 1 = 0$

Final Answer

The value of the integral is:

$\int_{-\pi}^{\pi} \sin(x) \, dx = 0$

Would you like any more details or have any questions about this solution?

Related Questions:

1. How would the integral change if the limits were from $0$ to $\pi$ instead of $-\pi$ to $\pi$?
2. Can you explain why the integral of an odd function over a symmetric interval is always zero?
3. What would be the integral of $\cos(x)$ over the same limits $[-π, π]$?
4. How would this problem be solved if we were dealing with a different trigonometric function, like $\sin^2(x)$?
5. Could you show how to solve the integral of $\sin(x)$ over a different interval, such as $[0, 2\pi]$?

Tip:

When dealing with integrals of trigonometric functions over symmetric intervals, it's useful to remember the properties of even and odd functions. This can often simplify the problem.