求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx

求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx
求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx

求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx
∫cos2x/[(sinx)^2*(cosx)^2]dx
=∫[(cosx)^2-(sinx)^2]/[(sinx)^2*(cosx)^2]dx
=∫[1/(sinx)^2-1/(cosx)^2]dx
=-cotx-tanx+c

∫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~