这道题利用分部积分
F(x) = ∫ (x * cos[2x]) dx = 1/2 * x * sin[2x] - 1/2 * ∫ sin[2x] dx
= 1/2 * x * sin[2x] + 1/4 * cos[2x] + C
∴ ∫ [0,π/2] (x * cos[2x]) dx = F(π/2]) - F(0) = -1/2.
计算定积分：∫n/2 n xcos2xdx
这道题利用分部积分
F(x) = ∫ (x * cos[2x]) dx = 1/2 * x * sin[2x] - 1/2 * ∫ sin[2x] dx
= 1/2 * x * sin[2x] + 1/4 * cos[2x] + C
∴ ∫ [0,π/2] (x * cos[2x]) dx = F(π/2]) - F(0) = -1/2.