1/cosx的原函数是ln|secx+tanx|+C。解答如下:
先算1/sinx原函数,S表示积分号
S1/sinxdx
=S1/(2sin(x/2)cos(x/2))dx
=S1/[tan(x/2)cos²(x/2)]d(x/2)
=S1/[tan(x/2)]d(tan(x/2))
=ln|zhitan(x/2)|+C
因为tan(x/2)=sin(x/2)/cos(x/2)=2sin²(x/2)/[2sin(x/2)cos(x/2)]=(1-cosx0/sinx=cscx-cotx
所以S1/sinxdx=ln|cscx-cotx|+C
S1/cosxdx
=S1/sin(x+派/2)d(x+派/2)
=ln|csc(x+派/2)-cot(x+派/2)|+C
=ln|secx+tanx|+C