1. **State the problem:** We need to evaluate the definite integral $$\int_0^{\frac{\pi}{2}} x \sin x \, dx$$. 2. **Formula and method:** To solve this integral, we use integration by parts. The formula for integration by parts is: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x \implies du = dx$$ $$dv = \sin x \, dx \implies v = -\cos x$$ 4. **Apply integration by parts:** $$\int_0^{\frac{\pi}{2}} x \sin x \, dx = \left. -x \cos x \right|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx$$ 5. **Evaluate the remaining integral:** $$\int_0^{\frac{\pi}{2}} \cos x \, dx = \left. \sin x \right|_0^{\frac{\pi}{2}} = 1 - 0 = 1$$ 6. **Evaluate the boundary term:** $$-x \cos x \Big|_0^{\frac{\pi}{2}} = -\frac{\pi}{2} \cdot \cos \frac{\pi}{2} + 0 \cdot \cos 0 = 0 + 0 = 0$$ 7. **Combine results:** $$\int_0^{\frac{\pi}{2}} x \sin x \, dx = 0 + 1 = 1$$ **Final answer:** $$\boxed{1}$$