# How do you integrate int 1/sqrt(x^2-6)  using trigonometric substitution?

Jan 25, 2016

$\int \setminus \frac{1}{\setminus \sqrt{{x}^{2} - 6}} = {\log}_{e} \left(\frac{x + \sqrt{{x}^{2} - 6}}{\setminus} \sqrt{6}\right) + c$

#### Explanation:

Given equation is $\int \setminus \frac{1}{\setminus \sqrt{{x}^{2} - 6}}$

For this equation, we'll substitute $x = \setminus \sqrt{6} \sec \left(t\right)$
So, $\mathrm{dx} = \setminus \sqrt{6} \sec \left(t\right) \tan \left(t\right) \mathrm{dt}$

Substituting these values into the equation, we get
$\int \setminus \frac{\setminus \sqrt{6} \sec \left(t\right) \tan \left(t\right) \mathrm{dt}}{\setminus \sqrt{6 {\sec}^{2} \left(x\right) - 6}}$
Taking $\setminus \sqrt{6}$ common from the denominator, we end with
$\int \setminus \frac{\sec \left(t\right) \tan \left(t\right) \mathrm{dt}}{\setminus \sqrt{{\sec}^{2} \left(t\right) - 1}}$

I believe you're well friends with this general trigonometric equation
${\tan}^{2} \left(\setminus \theta\right) + 1 = {\sec}^{2} \left(\setminus \theta\right)$
So subtracting $1$ from both sides, we get ${\tan}^{2} \left(\setminus \theta\right) = {\sec}^{2} \left(\setminus \theta\right) - 1$

Therefore, the denominator of the integral becomes
$\int \setminus \frac{\sec \left(t\right) \tan \left(t\right) \mathrm{dt}}{\tan \left(t\right)}$
Cancelling $\tan \left(t\right)$ function, we end with
$\int \sec \left(t\right) \mathrm{dt} = {\log}_{e} \left(\sec \left(t\right) + \tan \left(t\right)\right) + c$

Now, we know that $x = \setminus \sqrt{6} \sec \left(t\right) \setminus \implies \sec \left(t\right) = \frac{x}{\setminus} \sqrt{6}$
Also, from the above trigonometric identity, ${\tan}^{2} \left(t\right) = {x}^{2} / 6 - 1 = \frac{{x}^{2} - 6}{6} \setminus \implies \tan \left(t\right) = \setminus \sqrt{\setminus \frac{{x}^{2} - 6}{6}}$

So substituting the above values for their respective functions, we get the integral of the said main function.