8. (10.0分) 求定积分 int_(0)^pi x sin x dx.

设 $u = x$，$dv = \sin x \, dx$，则 $du = dx$，$v = -\cos x$。应用分部积分法： \[ \int_{0}^{\pi} x \sin x \, dx = \left[ -x \cos x \right]_{0}^{\pi} + \int_{0}^{\pi} \cos x \, dx = \pi + 0 = \pi \] 或令 $t = \pi - x$，则 \[ \int_{0}^{\pi} x \sin x \, dx = \int_{0}^{\pi} (\pi - t) \sin t \, dt = \pi \int_{0}^{\pi} \sin t \, dt - \int_{0}^{\pi} t \sin t \, dt = 2\pi - I \implies I = \pi \] 答案：$\boxed{\pi}$