What are the antiderivatives of #sec(x)#, #csc(x)# and #cot(x)#?

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Answer 1 · 2014-10-16T16:54:49

Since

#(ln|secx+tanx|)'={secxtanx+sec^2x}/{sec x+tanx}=secx#,

we have

#int secx dx=ln|secx+tanx|+C#

Since

#(-ln|cscx+cotx|)'=-{-cscxcotx-csc^2x}/{cscx+cotx}=cscx#,

we have

#int cscx dx=-ln|cscx+cotx|+C#

#int cotx dx=int{cosx}/{sinx}dx=ln|sinx|+C#

I hope that this was helpful.