# How do you integrate arcsec(x)?

Mar 9, 2015

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

You'll also need $\int \sec u \mathrm{du}$, 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:$\int a r c \sec \left(x\right) \mathrm{dx}$

Let $y = a r c \sec \left(x\right)$, so $x = \sec y$ and $\mathrm{dx} = \sec y \tan y \mathrm{dy}$.

With this substitution, the integral becomes:

$\int y \sec y \tan y \mathrm{dy}$.

Integrate this by parts:
Let $u = y$ and $\mathrm{dv} = \sec y \tan y \mathrm{dy}$.
Then $\mathrm{du} = \mathrm{dy}$ and $v = \sec y$.

$u v - \int v \mathrm{du} = y \sec y - \int \sec y \mathrm{dy}$
$= y \sec y - \ln \left\mid \sec y + \tan y \right\mid + C$.

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

The integral becomes:

$\int a r c \sec x \mathrm{dx} = \left(a r c \sec \left(x\right)\right) x - \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid + C$.

This is more easily read is we write it as:

$x \left(a r c \sec \left(x\right)\right) - \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid + C$.