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

$-cosx + c$

$c in RR$

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

$=> int sinx/ cosx * cosx dx$

$=> int sinx dx$

$=> -cosx + c$

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