Question

求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx

Answer 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~

追问

还真不懂~

追答

哪一步?不要说是全部0 0...
还有..答案错了,应该是-2/sin2x+C(常数)

Answer 2

∫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

追问

=∫1/(sinx)^2-1/(cosx)^2dx这没看懂

追答

根据积分公式得;
∫1/(sinx)^2 dx = -cotx +c
∫1/(cosx)^2 dx = tanx+c

∫[(cosx)^2-(sinx)^2]/[(sinx)^2*(cosx)^2]dx
=∫(cosx)^2/[(sinx)^2*(cosx)^2]-(sinx)^2/[(sinx)^2*(cosx)^2] dx (分子分母相约得)
=∫[1/(sinx)^2-1/(cosx)^2]dx