What is the integral of $sec (x)$ $sec(x)$ ?

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What is the integral of $sec (x)$ $sec(x)$ ?

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Answer 1 · 2018-04-18T18:07:53

$\int sec x d x = ln | sec x + tan x | + C$ $intsecxdx=ln|secx+tanx|+C$

Explanation:

Integrating the secant requires a bit of manipulation.

Multiply $sec x$ $secx$ by $\frac{sec x + tan x}{sec x + tan x}$ $(secx+tanx)/(secx+tanx)$ , which is really the same as multiplying by $1 .$ $1.$ Thus, we have

$\int (\frac{sec x (sec x + tan x)}{sec x + tan x}) d x$ $int((secx(secx+tanx))/(secx+tanx))dx$

$\int \frac{{sec}^{2} x + sec x tan x}{sec x + tan x} d x$ $int(sec^2x+secxtanx)/(secx+tanx)dx$

Now, make the following substitution:

$u = sec x + tan x$ $u=secx+tanx$

$d u = (sec x tan x + {sec}^{2} x) d x = ({sec}^{2} x + sec x tan x) d x$ $du=(secxtanx+sec^2x)dx=(sec^2x+secxtanx)dx$

We see that $d u$ $du$ appears in the numerator of the integral, so we may apply the substitution:

$\int \frac{d u}{u} = ln | u | + C$ $int(du)/u=ln|u|+C$

Rewrite in terms of $x$ $x$ to get

$\int sec x d x = ln | sec x + tan x | + C$ $intsecxdx=ln|secx+tanx|+C$

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