How do you integrate $arcsec(x)$ ?

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Answer 1 · 2015-03-09T13:50:13

Method: To integrate $arc sec (x)$ , substitution, then integrate by parts.

You'll also need $int secu du$ , which can be done by substitution and partial fractions.
Here's a nice explanation: http://socratic.org/questions/what-is-the-integral-of-sec-x .

Details: $int arcsec(x) dx$

Let $y=arc sec(x)$ , so $x=secy$ and $dx = secy tany dy$ .

With this substitution, the integral becomes:

$inty secy tany dy$ .

Integrate this by parts:
Let $u=y$ and $dv=secytany dy$ .
Then $du=dy$ and $v=secy$ .

$u v - int v du=ysecy-intsecy dy$
$=ysecy-ln abs(secy+tany)+C$ .

With $y=arc sec(x)$ we get $x=secy$ , and $tany=sqrt (x^2-1)$ .

The integral becomes:

$int arcsec x dx = (arc sec(x))x-ln abs(x+sqrt(x^2-1))+C$ .

This is more easily read is we write it as:

$x (arc sec(x)) - ln abs(x+sqrt(x^2-1))+C$ .

How do you integrate arcsec(x)arcsec(x)?