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1/cosx的原函数是多少

发布者:丁同华
导读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/

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