How do you integrate $a r c sec (x)$ $arcsec(x)$ ?

Question

33844 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-09T13:50:13

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

You'll also need $\int sec u d u$ $int secu du$ , 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: $\int a r c sec (x) d x$ $int arcsec(x) dx$

Let $y = a r c sec (x)$ $y=arc sec(x)$ , so $x = sec y$ $x=secy$ and $d x = sec y tan y d y$ $dx = secy tany dy$ .

With this substitution, the integral becomes:

$\int y sec y tan y d y$ $inty secy tany dy$ .

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

$u v - \int v d u = y sec y - \int sec y d y$ $u v - int v du=ysecy-intsecy dy$
$= y sec y - ln | sec y + tan y | + C$ $=ysecy-ln abs(secy+tany)+C$ .

With $y = a r c sec (x)$ $y=arc sec(x)$ we get $x = sec y$ $x=secy$ , and $tan y = \sqrt{x^{2} - 1}$ $tany=sqrt (x^2-1)$ .

The integral becomes:

$\int a r c sec x d x = (a r c sec (x)) x - ln ∣ ∣ x + \sqrt{x^{2} - 1} ∣ ∣ + C$ $int arcsec x dx = (arc sec(x))x-ln abs(x+sqrt(x^2-1))+C$ .

This is more easily read is we write it as:

$x (a r c sec (x)) - ln ∣ ∣ x + \sqrt{x^{2} - 1} ∣ ∣ + C$ $x (arc sec(x)) - ln abs(x+sqrt(x^2-1))+C$ .

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