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

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Answer 1 · 2018-05-20T13:09:00

The answer is #=ln(|tanx+secx|)-ln(|cosx|)+C#

We need

#intsecxdx=int((secx(tanx+secx)dx)/(tanx+secx))#

#=int((secxtanx+sec^2x)dx)/(tanx+secx)#

Let #u=tanx+secx#, #=>#, #du=(sec^2x+secxtanx)dx#

#intsecxdx=int(du)/u=ln(u)#

#=ln(|tanx+secx|)+C#

#inttanxdx=int(sinxdx)/cosx#

Let #u=cosx#, #=>#, #du=-sinxdx#

#inttanxdx=-int(du)/u=-ln(u)#

#=-ln(|cosx|)+C#

Therefore, the integral is

#I=int(secx+tanx)dx#

#=intsecxdx+inttanxdx#

#=ln(|tanx+secx|)-ln(|cosx|)+C#