计算定积分∫_0^(π/2)▒〖x(sin⁡x+cos⁡x 〗)dx.小女子

先算不定积分,需要用分部积分法
∫ x(sinx+cosx ）dx = ∫ x d(-cosx + sinx) = x(sinx - cosx) - ∫ (sinx - cosx) dx
= x(sinx - cosx) - (-cosx - sinx) + C = x(sinx - cosx) + (sinx + cosx) + C
再代入定积分的上下标
原式 = [π/2 (1 - 0) + (1 + 0) ] - [0 (0 - 1) + (0 + 1)] = π/2 + 1 - 1 = π/2