# How do you integrate intxsec^-1(x)dx?

Apr 10, 2018

The answer is $= {x}^{2} / 2 a r c \sec x - \frac{1}{2} \sqrt{{x}^{2} - 1} + C$

#### Explanation:

Perform the integration by parts

$\int u v ' = u v - \int u ' v$

Here,

$u = a r c \sec x$, $\implies$, $u ' = \frac{\mathrm{dx}}{x \sqrt{{x}^{2} - 1}}$

$v ' = x$, $\implies$, $v = {x}^{2} / 2$

Therefore,

$\int x a r c \sec x \mathrm{dx} = {x}^{2} / 2 a r c \sec x - \frac{1}{2} \int \frac{x \mathrm{dx}}{\sqrt{{x}^{2} - 1}}$

Let $u = {x}^{2} - 1$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

Therefore,

$\frac{1}{2} \int \frac{x \mathrm{dx}}{\sqrt{{x}^{2} - 1}} = \frac{1}{4} \int \frac{\mathrm{du}}{\sqrt{u}}$

$= \frac{1}{2} \sqrt{u}$

$= \frac{1}{2} \sqrt{{x}^{2} - 1}$

And finally,

$\int x a r c \sec x \mathrm{dx} = {x}^{2} / 2 a r c \sec x - \frac{1}{2} \sqrt{{x}^{2} - 1} + C$