We are asked to compute the integral of $x \sin(x)$ from $-\pi$ to $\pi$:

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

Step 1: Recognize the function's properties

The function $x \sin(x)$ is an odd function because:

• $x$ is odd ($f(-x) = -f(x)$),
• $\sin(x)$ is odd ($\sin(-x) = -\sin(x)$),
• the product of two odd functions is an even function.

Thus, $x \sin(x)$ is an odd function.

Step 2: Use symmetry of definite integrals

The integral of any odd function over a symmetric interval $[-a, a]$ is zero. In this case, since the interval is $[- \pi, \pi]$, and $x \sin(x)$ is odd, we can directly conclude:

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

Final Answer:

$I = 0$

Would you like a detailed breakdown of the integration process or more examples?

Here are some related questions for you:

1. How would you integrate $\int_{-\pi}^{\pi} \cos(x) \, dx$?
2. What is the general property of definite integrals for odd and even functions?
3. How do we handle integrals of mixed functions (e.g., $x \cos(x)$)?
4. What is the concept of symmetry in definite integrals?
5. How do you calculate the area under a curve for non-symmetric intervals?

Tip: Always check the symmetry of the function when integrating over symmetric intervals, as it can save a lot of computation!