# How do you integrate int (dx) / ( sqrt(x^(2) - 1 )  from -2 to -3?

Apr 20, 2018

${\int}_{-} {3}^{-} 2 \frac{1}{\sqrt{{x}^{2} - 1}} \mathrm{dx} \approx 0.44578927712$

#### Explanation:

We have:

${\int}_{-} {3}^{-} 2 \frac{1}{\sqrt{{x}^{2} - 1}} \mathrm{dx}$

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$ when $F ' \left(x\right) = f \left(x\right)$

What is $\int \frac{1}{\sqrt{{x}^{2} - 1}} \mathrm{dx}$?

We use the trigonometric substitution.

Since the variable is getting subtracted by one, this is the secant case.

We draw a right triangle:

We see that:

$\sec \left(\theta\right) = x$

$\implies \theta = a r c \sec \left(x\right)$

$\implies \sec \left(\theta\right) \tan \left(\theta\right) d \theta = \mathrm{dx}$

$\tan \left(\theta\right) = \frac{\sqrt{{x}^{2} - 1}}{1}$

$\implies \tan \left(\theta\right) = \sqrt{{x}^{2} - 1}$

Substitute.

$\implies \int \frac{1}{\tan \left(\theta\right)} \sec \left(\theta\right) \tan \left(\theta\right) d \theta$ Simplify

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

This is one of the "basic" integrals you should memorize.

$\implies \ln \left\mid \sec \left(\theta\right) + \tan \left(\theta\right) \right\mid$ substitute

$\implies \ln \left\mid \sec \left(a r c \sec \left(x\right)\right) + \tan \left(a r c \sec \left(x\right)\right) \right\mid$

$\implies \ln \left\mid x + \tan \left(a r c \sec \left(x\right)\right) \right\mid$

Therefore:

${\int}_{-} {3}^{-} 2 \frac{1}{\sqrt{{x}^{2} - 1}} \mathrm{dx} = {\left[\ln \left\mid x + \tan \left(a r c \sec \left(x\right)\right) \right\mid\right]}_{-} {3}^{-} 2$

$\implies \ln \left\mid - 3 + \tan \left(a r c \sec \left(- 3\right)\right) \right\mid - \ln \left\mid - 2 + \tan \left(a r c \sec \left(- 2\right)\right) \right\mid$

$\implies 1.76274717404 - 1.31695789692$

$\implies 0.44578927712$