# How do you integrate int dx/sqrt(x^2-64) using trig substitutions?

Jun 28, 2018

$I = \ln | \sqrt{{x}^{2} / 64 - 1} + \frac{x}{8} | + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{{x}^{2} - 64}} \setminus \mathrm{dx}$

We can perform the substitution:

$x = 8 \sec \theta \implies \sec \theta = \frac{x}{8}$

And differentiating wrt $\theta$:

$\frac{\mathrm{dx}}{d \theta} = 8 \sec \theta \tan \theta$

And Substituting into the integral, it becomes:

$I = \int \setminus \frac{1}{\sqrt{{\left(8 \sec \theta\right)}^{2} - 64}} \setminus 8 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus = \int \setminus \frac{8 \sec \theta \tan \theta}{\sqrt{64 {\sec}^{2} \theta - 64}} \setminus d \theta$

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

$\setminus \setminus = \int \setminus \frac{\sec \theta \tan \theta}{\sqrt{{\tan}^{2} \theta}} \setminus d \theta$

$\setminus \setminus = \int \setminus \sec \theta \setminus d \theta$

$\setminus \setminus = \ln | \tan \theta + \sec \theta | + C$

And using the identity:

$1 + {\tan}^{2} A \equiv {\sec}^{2} A \implies \tan \theta = \sqrt{{\sec}^{2} \theta - 1}$

Allowing us to restore the earlier substitution:

$I = \ln | \sqrt{{\sec}^{2} \theta - 1} + \sec \theta | + C$

$\setminus \setminus = \ln | \sqrt{{\left(\frac{x}{8}\right)}^{2} - 1} + \frac{x}{8} | + C$

$\setminus \setminus = \ln | \sqrt{{x}^{2} / 64 - 1} + \frac{x}{8} | + C$