What's the integral of #int (tanx/secx) dx#?

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Answer 1 · 2018-05-31T16:21:31

The answer is #=-cosx+C#

Perform this integral by substitution

Let #u=1/secx#

#=>#, #du=-(secxtanx)/sec^2xdx=-tanx/secxdx#

The integral is

#I=int(tanxdx)/(secx)=int(-du)#

#=-u#

#=-1/secx#

#=-cosx+C#

Answer 2 · 2018-05-31T17:42:21

#-cosx + c #

#c in RR #

#=> int sinx / cosx * 1/ secx dx #

#=> int sinx/ cosx * cosx dx #

#=> int sinx dx #

#=> -cosx + c #