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

$I=int(secx+tanx)dx$

$=intsecxdx+inttanxdx$

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

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