Question

求原函数

(Cosx)^-3的原函数。谢谢～

Answer 1

f1/cos^3xdx
先算f1/sin^3xdx
=f1/{[2tan(x/2)/[1+tan(x/2)^2]}^3dx
=fsec^6(x/2)/[8tan^3(x/2)]dx
=f[sec^4(x/2)sec^2(x/2)]/[8tan^3(x/2)]dx
设:tan(x/2)=u 则：1/2sec^2(x/2)dx=du sec^2(x/2)dx=2du
=f[(u^2+1)^2]/[4u^3]du
=1/4f [u^4+2u^2+1]/u^3du
=1/4[ u+2/u+u^(-3)]du
=1/4[1/2u^2+2lnu-1/2u^(-2)]+c
=1/8[u^2-u^(-2)]+1/2lnu+c
=1/8[tan^2(x/2)-cot^2(x/2)]+1/2ln|tan(x/2)|+c
=1/2[ln|tan(x/2)|-cotxcscx]+c......................1
再算：
f1/cos^3xdx
=f1/sin^3(x+pai/2)dx
=f1/sin^3(x+pai/2)d(x+pai/2)
将1式代入：
=1/2[ln|tan(x/2+pai/4)|-cot(x+pai/2)csc(x+pai/2)]+c
=1/2[ln|tan(x/2+pai/4)|)+tanxsecx]+C