Question

求不定积分：1/（sinx）^3

不查表怎么去求它的原函数啊

Answer 1

∫1/（sinx）^3dx=∫sinx/（sinx）^4dx
=∫sinx/（1-cosx^2）^2dx
=-∫1/(1-cosx^2）^2dcosx
1/(1-u^2）^2=A/(1-u)^2+B/(1+u)^2+C/(1-u)+D/(1+u)
A=1/4,B=1/4,C=1/4,D=1/4;
∫1/(1-u^2）^2du=1/4[(/1-u)-(1/1+u)-ln|1-u|+ln|1+u|]+C( |u|<1)
将COSx代入即得

Answer 2

解：
∫1/（sinx）^3dx
=∫1/(sinx(1-(cosx)^2)）dx
=(1/2)∫(1/sinx)[1/(1+cosx)+1/(1-cosx)]dx
=(1/2)∫sinx/(sinxsinx(1+cosx))dx+(1/2)∫sinx/(sinxsinx(1-cosx))dx
=(1/2)∫sinx/(1-(cosx)^2)(1+cosx))dx+(1/2)∫sinx/((1-(cosx)^2)(1-cosx))dx
=-(1/2)∫1/((1-(cosx)^2)(1+cosx))d(cosx)-(1/2)∫1/((1-(cosx)^2)(1-cosx))d(cosx)
=(-1/2)∫(1/2)(1/(1+cosx)+1/(1-cosx))(1/(1+cosx))dcosx
+(-1/2)∫(1/2)(1/(1+cosx)+1/(1-cosx))(1/(1-cosx))dcosx
=(1/4)(1/(1+cosx))-1/8ln(1+cosx)+(1/8)ln(1-cosx)-(1/4)1/(1-cosx)
-1/8ln(1+cosx)+(1/8)ln(1-cosx)
=-cosx/(sinx)^2/2+ln((1-cosx)/sinx)/2+C