# How do you integrate int dx/sqrt(x^2-4) using trig substitutions?

Apr 6, 2018

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

#### Explanation:

If our integral involves a root in the form of $\sqrt{{x}^{2} - {a}^{2}} ,$ we're best off using the substitution

$x = a \sec \theta .$ Here, we see ${a}^{2} = 4 , a = 2 ,$ so we use the substitution

$x = 2 \sec \theta$

$\mathrm{dx} = 2 \sec \theta \tan \theta d \theta$

Apply to the integral $\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 4}} ,$ yielding

$\int \frac{2 \sec \theta \tan \theta}{\sqrt{4 {\sec}^{2} \theta - 4}} d \theta$

Simplify.

int(2secthetatantheta)/sqrt(4(sec^2theta-1)d theta

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

Recalling the identity ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta ,$ we get

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

So, we end up with the common integral (you should have this memorized)

$\int \sec \theta d \theta = \ln | \sec \theta + \tan \theta | + C$

We need things in terms of $x .$ Well, recalling that $x = 2 \sec \theta , \sec \theta = \frac{x}{2.}$ We still need to find tangent.

Recalling that ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta , {x}^{2} / 4 - \frac{4}{4} = {\tan}^{2} \theta , {\tan}^{2} \theta = \frac{{x}^{2} - 4}{4} , \tan \theta = \frac{\sqrt{{x}^{2} - 4}}{2}$

Thus, we have

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

$\ln | \frac{x + \sqrt{{x}^{4} - 4}}{2} | = \ln | x + \sqrt{{x}^{2} - 4} | - \ln | 2 | + C = \ln | x + \sqrt{{x}^{2} - 4} | + C$, as we can absorb that logarithm into the constant of integration.

Finally,

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

Apr 6, 2018