# How do you integrate int 1/(xsqrt(x^2-1) )dx using trigonometric substitution?

Nov 5, 2017

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

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{x \sqrt{{x}^{2} - 1}} \setminus \mathrm{dx}$

Let us attempt a substitution of the form:

$\sec \theta = x$

Then differentiating wrt $x$ we have:

$\sec \theta \tan \theta \frac{d \theta}{\mathrm{dx}} = 1$

Substituting into the integral we have:

$I = \int \setminus \frac{1}{\sec \theta \sqrt{{\sec}^{2} \theta - 1}} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{1}{\sec \theta \sqrt{{\tan}^{2} \theta}} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{1}{\sec \theta \tan \theta} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus d \theta$
$\setminus \setminus = \theta + C$

Restoring the substitution gives:

$I = a r c \sec x + C$