# Using int sec x dx = ln|sec x + tan x | + c , find int 1/ sqrt (2x^2 - 4) dx using suitable trigonometric substitution?

Jul 18, 2017

$\int \setminus \frac{1}{\sqrt{2 {x}^{2} - 4}} \setminus \mathrm{dx} = \frac{\sqrt{2}}{2} \setminus \ln | x + \sqrt{\left({x}^{2} - 2\right)} | + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{2 {x}^{2} - 4}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{4 \left(\frac{1}{2} {x}^{2} - 1\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{\sqrt{{\left(\frac{x}{\sqrt{2}}\right)}^{2} - 1}} \setminus \mathrm{dx}$

Comparing with the trig identity ${\tan}^{2} A \equiv {\sec}^{2} A - 1$, Consider a substitution of the form:

$\frac{x}{\sqrt{2}} = \sec \theta \implies \frac{\mathrm{dx}}{d \theta} = \sqrt{2} \sec \theta \tan \theta$

If we substitute this into our integral we have:

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

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

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

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

$\setminus \setminus = \frac{\sqrt{2}}{2} \setminus \ln | \sec \theta + \tan \theta | + A$

Again Using ${\tan}^{2} A \equiv {\sec}^{2} A - 1$ we have:

${\tan}^{2} \theta \equiv {x}^{2} / 2 - 1 = \frac{{x}^{2} - 2}{2}$
$\implies \tan \theta = \sqrt{\frac{{x}^{2} - 2}{2}}$

And using the above results to restore the substitution we have:

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