请问1⼀cosx的原函数怎么算,要过程哦

谢谢是求原函数,不是反函数啊
2025-06-20 07:16:32
推荐回答(3个)
回答1:

先算一下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|tan(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

回答2:

∫1/cosxdx
=∫secx(secx+tanx)/(secx+tanx)dx
=∫1/(secx+tanx)d(secx+tanx)
=ln/secx+tanx/+C

/secx+tanx/是secx+tanx的绝对值

回答3:

y=1/cosx
x=arccos(1/y)