# How do you evaluate int dx/(x^2sqrt(x^2 - 9)) with x = 3sec(theta)?

## Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.) $\int$ $\frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}}$, x = 3sec($\theta$)

Jul 5, 2018

$I = \int$ $\frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 9}}$

Let $x = 3 \sec \theta$

$\implies \mathrm{dx} = 3 \sec \theta \tan \theta d \theta$

So
$I = \int \frac{3 \sec \theta \tan \theta d \theta}{9 {\sec}^{2} \theta \sqrt{9 {\sec}^{2} \theta - 9}}$

$= \int \frac{3 \sec \theta \tan \theta d \theta}{27 {\sec}^{2} \theta \sqrt{{\sec}^{2} \theta - 1}}$

$= \frac{1}{9} \int \frac{\sec \theta \tan \theta d \theta}{{\sec}^{2} \theta \tan \theta}$

$= \frac{1}{9} \int \cos \theta d \theta$

$= \frac{1}{9} \sin \theta + C$

$= \frac{1}{9} \tan \frac{\theta}{\sec} \theta + C$

$= \frac{1}{9} \frac{\sqrt{{\sec}^{2} \theta - 1}}{\sec} \theta + C$

$= \frac{1}{9} \frac{\sqrt{{x}^{2} / 9 - 1}}{\frac{x}{3}} + C$

$= \frac{1}{9} \frac{\sqrt{{x}^{2} - 9}}{x} + C$