# How do you integrate int 1/(x^2sqrt(x^2-7)) by trigonometric substitution?

Nov 25, 2016

$\frac{\sqrt{{x}^{2} - 7}}{7 x} + C$

#### Explanation:

$I = \int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 7}}$

Apply the substitution $x = \sqrt{7} \sec \theta , \mathrm{dx} = \sqrt{7} \sec \theta \tan \theta d \theta$.

$\textcolor{w h i t e}{I} = \int \frac{\sqrt{7} \sec \theta \tan \theta d \theta}{7 {\sec}^{2} \theta \sqrt{7 {\sec}^{2} \theta - 7}}$

$\textcolor{w h i t e}{I} = \frac{1}{7} \int \frac{\tan \theta d \theta}{\sec \theta \sqrt{{\sec}^{2} \theta - 1}}$

$\textcolor{w h i t e}{I} = \frac{1}{7} \int \frac{d \theta}{\sec} \theta$

$\textcolor{w h i t e}{I} = \frac{1}{7} \int \cos \theta d \theta$

$\textcolor{w h i t e}{I} = \frac{1}{7} \sin \theta$

$\textcolor{w h i t e}{I} = \frac{1}{7} \tan \frac{\theta}{\sec} \theta$

$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\sqrt{{\sec}^{2} \theta - 1}}{\sec} \theta$

Using $x = \sqrt{7} \sec \theta \implies \sec \theta = \frac{x}{\sqrt{7}}$:

$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\sqrt{{x}^{2} / 7 - 1}}{\frac{x}{\sqrt{7}}}$

$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\frac{1}{\sqrt{7}} \sqrt{{x}^{2} - 7}}{\frac{x}{\sqrt{7}}}$

$\textcolor{w h i t e}{I} = \frac{\sqrt{{x}^{2} - 7}}{7 x} + C$