第4个回答 2018-12-01
u=cosx
∫ dx/[ sinx.(cosx)^3]
=-∫ dcosx/[ (sinx)^2.(cosx)^3]
=-∫ du/[ u^3.(1-u^2)]
=-∫ { 1/u +1/u^3 + (1/2)[1/(1-u)] -(1/2) [1/(1+u)] } du
=- [ ln|u| - (1/2)(1/u^2) + (1/2) ln|1/(1-u^2) | ] +C
= [ -ln|u| + (1/2)(1/u^2) + (1/2) ln|1-u^2 | ] +C
= [ -ln|cosu| + (1/2)(1/(cosu)^2) + ln|sinu | ] +C
let
1/[ u^3.(1-u^2)] ≡ A/u +B/u^2 +C/u^3 + D/(1-u) +F/(1+u)
=>
1≡ Au^2.(1-u^2) +Bu(1-u^2) +C(1-u^2) + Du^3.(1+u) +Fu^3.(1-u)
u=0, => C= 1
u=1, => D = 1/2
u=-1, => F=-1/2
coef. of u^4
-A+D-F=0
-A+1=0
A=1
coef. of u^3
-B+D+F=0
B=0
1/[ u^3.(1-u^2)]
≡ 1/u +1/u^3 + (1/2)[1/(1-u)] -(1/2) [1/(1+u)]