1. **State the problem:** Evaluate the definite integral $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x \, dx$$. 2. **Recall the formula and properties:** The integral of a product involving an odd function and an even function over symmetric limits can be simplified by analyzing the parity of the integrand. 3. **Check the parity of the integrand:** - $x$ is an odd function since $x(-t) = -x(t)$. - $\sin x$ is an odd function since $\sin(-x) = -\sin x$. - The product $x \sin x$ is an even function because the product of two odd functions is even: $$f(-x) = (-x) \sin(-x) = (-x)(-\sin x) = x \sin x = f(x)$$. 4. **Use the property of even functions:** For an even function $f(x)$, $$\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx$$. 5. **Rewrite the integral:** $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x \, dx = 2 \int_0^{\frac{\pi}{2}} x \sin x \, dx$$. 6. **Use integration by parts:** Let - $u = x \implies du = dx$ - $dv = \sin x \, dx \implies v = -\cos x$ 7. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 8. **Calculate:** $$\int_0^{\frac{\pi}{2}} x \sin x \, dx = \left[-x \cos x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx$$ 9. **Evaluate the boundary term:** $$-x \cos x \bigg|_0^{\frac{\pi}{2}} = -\frac{\pi}{2} \cos \frac{\pi}{2} + 0 = 0$$ 10. **Evaluate the remaining integral:** $$\int_0^{\frac{\pi}{2}} \cos x \, dx = \sin x \bigg|_0^{\frac{\pi}{2}} = 1 - 0 = 1$$ 11. **Combine results:** $$\int_0^{\frac{\pi}{2}} x \sin x \, dx = 0 + 1 = 1$$ 12. **Multiply by 2 for the original integral:** $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x \, dx = 2 \times 1 = 2$$ **Final answer:** $$\boxed{2}$$