Question

对1/(1-t^2)^2求原函数

Answer 1

∫dt/(1-t^2)^2
let
t= siny
dt =cosydy

∫dt/(1-t^2)^2
=∫dy/(cosy)^3
=∫(secy)^3dy

consider
∫(secy)^3dy = ∫secydtany
=secy. tany - ∫secy.(tany)^2dy
=secy. tany - ∫secy.[(secy)^2-1]dy
2∫(secy)^3dy =secy. tany + ∫secydy
=secy. tany + ln|secy+tany|
∫(secy)^3dy = (1/2)[secy. tany + ln|secy+tany| ] + C
=(1/2)[secy. tany + ln|secy+tany| ] + C
= (1/2)[ t/(1-t^2) + ln|√(1-t^2) + t/√(1-t^2) ] + C

∫dt/(1-t^2)^2
=∫(secy)^3dy

=(1/2)[ t/(1-t^2) + ln|√(1-t^2) + t/√(1-t^2) ] + C

Answer 2

换元法，令t=cosx
dt=-sinxdx
代入就可以了

追问

这个问题就是从求1/cosx^3原函数变到这一步的，别再变回去 了