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

Jun 28, 2016

Use x = 2*sec(u) as a first substitution.

#### Explanation:

With $x = 2 \cdot \sec \left(u\right) , \therefore \mathrm{dx} = 2 \cdot \sec \left(u\right) \cdot \tan \left(u\right)$

$\sqrt{{x}^{2} - 4} = \sqrt{4 \cdot {\sec}^{2} \left(u\right) - 4} = 2 \cdot \sqrt{{\sec}^{2} \left(u\right) - 1}$

From ${\sin}^{2} \left(a\right) + {\cos}^{2} \left(a\right) = 1$, divide through by ${\cos}^{2} \left(a\right)$ to obtain

${\tan}^{2} \left(a\right) + 1 = {\sec}^{2} \left(a\right)$

So, we have:

int 2/(sqrt(x^2-4) dx  = int (4*sec(u)*tan(u))/(2sqrt(tan^2(u)) $\mathrm{du}$

So, integral of $2 \cdot \int \sec \left(u\right) \mathrm{du}$

Multiply numerator and denominator by $\sec \left(u\right) + \tan \left(u\right)$ to obtain

$2 \cdot \int \frac{{\sec}^{2} \left(u\right) + \sec \left(u\right) \cdot \tan \left(u\right)}{\sec \left(u\right) + \tan \left(u\right)} \mathrm{du}$

Use substitution $r = \sec \left(u\right) + \tan \left(u\right)$ which gives

$2 \cdot \int \frac{\mathrm{dr}}{r}$

$= 2 \cdot \ln \left(r\right) + C$

$= 2 \cdot \ln \left(\sec \left(u\right) + \tan \left(u\right)\right) + C$

 = 2*ln(sec(arcsec(x/2) + tan(arcsec(x/2)) + C

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

And there we have it, let me know if you need any help with the last bit but it's fairly easy to figure out if you look at a few triangles.