∫x(sinx)^2dx用分部积分法求

2025-03-10 16:13:35
推荐回答(2个)
回答1:

回答2:

∫x(sinx)^2dx
=(1/2)∫x(1-cos2x)dx
=(1/4)x^2-(1/2)∫xcos2xdx
=(1/4)x^2-(1/4)∫xdsin2x
=(1/4)x^2-(1/4)xsin2x +(1/4)∫sin2x dx
=(1/4)x^2-(1/4)xsin2x -(1/8)cos2x + C