求不定积分∫cos2x⼀[(sinx)^2(cosx)^2] dx

2024-11-05 02:27:09
推荐回答(1个)
回答(1):

∫cos2x/[(sinx)^2(cosx)^2]
dx
=1/2∫dsin2x/[(sinx)^2(cosx)^2]
=1/2∫dsin2x/[1/4(sin2x)^2]
=2∫dsin2x/(sin2x)^2
=-2/sin2x
这个方法应该叫凑微分~
不懂就追问哈~