# Evaluate the integral with hyperbolic or trigonometric substitution. ?

Mar 29, 2018

$\frac{\sqrt{6}}{144} \sin \left(2 {\sec}^{-} 1 \sqrt{6} \frac{x}{6}\right) + \frac{V}{72} {\sec}^{-} 1 \left(\sqrt{6} \frac{x}{6}\right) + C$

#### Explanation:

by applying the trigonometric substitution:
$x = \sqrt{6} \sec \left(u\right)$
$\mathrm{dx} = \sqrt{6} \sec \left(u\right) \cdot \tan \left(u\right) \mathrm{du}$
$\int \left(\frac{\sqrt{6} \sec u \cdot \tan u \mathrm{du}}{6 \cdot \sqrt{6} \cdot {\sec}^{2} u \cdot \sqrt{6} \tan u}\right)$

after simplification
$\frac{\sqrt{6}}{36} \int \left(\frac{1}{\sec} ^ 2 \left(u\right) \mathrm{du}\right)$=$\frac{\sqrt{6}}{36} \int \left({\cos}^{2} \left(u\right) \mathrm{du}\right)$

by using double-angle formulae: ${\cos}^{2} u = \frac{1}{2} \left(1 + \cos 2 u\right)$

$\frac{\sqrt{6}}{72} \int \left(1 + \cos 2 u\right) \mathrm{du}$

=$\frac{\sqrt{6}}{72} \left(u + \frac{1}{2} \sin 2 u\right)$
by substituting back $u = {\sec}^{-} 1 \left(\frac{x}{\sqrt{6}}\right)$
you get:
$\int \frac{\mathrm{dx}}{{x}^{3} \cdot \sqrt{{x}^{2} - 6}} = \frac{\sqrt{6}}{144} \sin \left(2 {\sec}^{-} 1 \sqrt{6} \frac{x}{6}\right) + \frac{V}{72} {\sec}^{-} 1 \left(\sqrt{6} \frac{x}{6}\right) + C$