# How do you integrate int 1/(sqrt(x^2-4))dx using trigonometric substitution?

Mar 8, 2016

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

#### Explanation:

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

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

Let $\frac{x}{2} = \sec \left(\theta\right)$
Then $\mathrm{dx} = 2 \sec \left(\theta\right) \tan \left(\theta\right)$

$\implies \int \frac{1}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = \frac{1}{2} \int \frac{1}{\sqrt{{\sec}^{2} \left(\theta\right) - 1}} \cdot 2 \sec \left(\theta\right) \tan \left(\theta\right) d \theta$

$= \int \frac{\sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{{\tan}^{2} \left(\theta\right)}} d \theta$

$= \int \sec \left(\theta\right) d \theta$

$= \ln | \sec \left(\theta\right) + \tan \left(\theta\right) | + C$

$= \ln | \frac{x}{2} + \sqrt{{x}^{2} / 4 - 1} | + C$

(For the final equality, try drawing a right triangle with angle $\theta$ such that $\sec \left(\theta\right) = \frac{x}{2}$ and solve for $\tan \left(\theta\right)$)

We could stop here, but we can also make this look a little nicer by getting rid of the $\frac{1}{2}$

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

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

But as $C$ is an arbitrary constant, we can include the $\ln \left(\frac{1}{2}\right)$ in that to give us the final result

$\ln | x + \sqrt{{x}^{2} - 4} | + C$