(x^2-1)cos2xdx,求不定积分
收藏:
0
点赞数:
0
评论数:
0
1个回答

(x^2-1)cos2xdx

=∫(x²-1)cos2xdx

=(1/2)∫(x²-1)d(sin2x)

=(1/2)[(x²-1)sin2x-∫2xsin2xdx]

=(1/2)[(x²-1)sin2x+∫xd(cos2x)]

=(1/2)[(x²-1)sin2x+xcos2x-∫cos2xdx]

=(1/2)[(x²-1)sin2x+xcos2x-(1/2)sin2x]+C

=(1/2)[(x²-3/2)sin2x+xcos2x]+C

点赞数:
0
评论数:
0
相关问题
关注公众号
一起学习,一起涨知识