不定积分分部积分法题目

2024-12-04 22:51:39
推荐回答(1个)
回答(1):

∫xtan^2xdx
= ∫x(sec^2x-1)dx
=∫xdtanx-∫xdx
=xtanx-∫tanxdx-(1/2)x^2
=xtanx+ln|cosx|-(1/2)x^2+c
∫xsinxcosxdx
= (1/2)∫xdsin2x
=(1/2)xsin2x-(1/2)∫sin2xdx
=(1/2)xsin2x+(1/4)cos2x+c.