To evaluate the integral

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \sin(x) \, dx,$

we can use symmetry properties of the sine function. Here's the process:

Step 1: Analyze the integrand

The integrand is $3 \sin(x)$. The sine function, $\sin(x)$, is odd, meaning that:

$\sin(-x) = -\sin(x).$

When multiplied by the constant $3$, the function $3\sin(x)$ is still odd.

Step 2: Use symmetry for odd functions

The integral of an odd function over a symmetric interval $[-a, a]$ is always $0$. Mathematically:

$\int_{-a}^a f(x) \, dx = 0 \quad \text{if } f(x) \text{ is odd}.$

In this case, $f(x) = 3 \sin(x)$ is odd, and the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is symmetric about $0$.

Step 3: Conclusion

By the symmetry property of odd functions, the integral evaluates to:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \sin(x) \, dx = 0.$

Would you like additional clarification or further examples?

Related Questions:

1. How does the symmetry property work for even functions?
2. What happens if the limits of integration are not symmetric?
3. Can you apply this symmetry rule for integrals with trigonometric functions other than sine?
4. How does the behavior of the constant $3$ affect the result of this integral?
5. What are some examples of real-world scenarios involving symmetric integrals?

Tip: Always identify whether the integrand is odd or even when working with symmetric limits of integration.