求它的递推公式！
2008-08-30 15:54:35晓***
求它的递推公式！：1.
I(n)=-∫d(cosx)/[sinx]^(n+1)=
=-cosx/[sinx]^(n+1)+∫cosxd[1/(sinx)^(?
最佳回答
1.
I(n)=-∫d(cosx)/[sinx]^(n+1)=
=-cosx/[sinx]^(n+1)+∫cosxd[1/(sinx)^(n+1)]=
=-cosx/[sinx]^(n+1)-(n+1)∫(cosx)^2dx/(sinx)^(n+2)=
=-cosx/[sinx]^(n+1)-(n+1)[I(n+2)-I(n)]
==>
I(n+2)=-cosx/[(n+1)(sinx)^(n+1)]+(nI(n))/(n+1)
2.
I(1)=(1/2)ln[(1-cosx)/(1+cosx)]+C
I(2)=-ctgx+C.
2008-09-01 13:26:49
很赞哦！ (262)
