# How do you find the antiderivative of int sqrt(x^2-1) dx?

Mar 5, 2017

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

#### Explanation:

When you say you want the "anti-derivative" of $\int \sqrt{{x}^{2} - 1} \mathrm{dx}$, I am assuming you mean you want the "anti-derivative" that is $\int \sqrt{{x}^{2} - 1} \mathrm{dx}$.

Otherwise you are looking for: $\int \left(\int \sqrt{{x}^{2} - 1} \mathrm{dx}\right) \setminus \mathrm{dx}$.

For $\int \sqrt{{x}^{2} - 1} \setminus \mathrm{dx}$, there is a convenient hyperbolic relationship:

${\cosh}^{2} z - {\sin}^{2} z = 1$

If we let $x = \cosh z , \mathrm{dx} = \sinh z \setminus \mathrm{dz}$, we get this:

$\int \sqrt{{\cosh}^{2} z - 1} \sinh z \setminus \mathrm{dz}$

$\int \sqrt{{\sinh}^{2} z} \sinh z \setminus \mathrm{dz}$

$= \int {\sinh}^{2} z \setminus \mathrm{dz}$

By the hyperbolic double angle formula ($\cosh 2 z = 1 + 2 {\sinh}^{2} z$):

$= \frac{1}{2} \int \cosh 2 z - 1 \setminus \mathrm{dz}$

$= \frac{1}{2} \left(\frac{1}{2} \sinh 2 z - z\right) + C = \frac{1}{4} \sinh 2 z - \frac{1}{2} z + C q \quad \triangle$

Now:

$\sinh 2 z = 2 \sinh z \cosh z =$

$= 2 \sqrt{{x}^{2} - 1} \cdot x$

So $\triangle$ becomes:

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

NB

If you like we can take ${\cosh}^{- 1} \left(x\right)$ and move it from hyperbolic to natural log status:

let $y = {\cosh}^{- 1} \left(x\right)$

$x = \cosh y$

$= \frac{{e}^{y} + {e}^{- y}}{2}$

$\implies x = \frac{{e}^{y} + {e}^{- y}}{2}$

$\implies 2 x {e}^{y} = {e}^{2 y} + 1$

$\implies {\left({e}^{y}\right)}^{2} - 2 x {e}^{y} + 1 = 0$

${e}^{y} = \frac{2 x \pm \sqrt{4 {x}^{2} - 4}}{2} = x \pm \sqrt{{x}^{2} - 1}$

$\implies y = \ln \left(x \pm \sqrt{{x}^{2} - 1}\right)$

So $\triangle$ becomes:

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

Mar 5, 2017

$\int \sqrt{{x}^{2} - 1}$ $\mathrm{dx} = \frac{1}{2} \left(x \sqrt{{x}^{2} - 1} - \ln | x + \sqrt{{x}^{2} - 1} |\right) + \text{c}$

#### Explanation:

We are trying to evaluate $\int \sqrt{{x}^{2} - 1}$ $\mathrm{dx}$.

Firstly, we should let $x = \sec u$

Then our integrand becomes $\sqrt{{\sec}^{2} u - 1} = \tan u$ and

$\frac{\mathrm{dx}}{\mathrm{du}} = \sec u \tan u \Rightarrow \mathrm{dx} = \sec u \tan u$ $\mathrm{du}$

$\int \sqrt{{x}^{2} - 1}$ $\mathrm{dx} = \int \sec u {\tan}^{2} u$ $\mathrm{du} =$
$\int \sec u \left({\sec}^{2} u - 1\right)$ $\mathrm{du} = \int {\sec}^{3} u - \sec u$ $\mathrm{du} =$
$\int {\sec}^{3} u$ $\mathrm{du} - \int \sec u$ $\mathrm{du}$

The integral of $\sec u$ is a standard and is evaluated as $\ln | \sec u + \tan u | + \text{C}$.

The integral of ${\sec}^{3} u$ will require integration by parts. If you wish to know how to do it, I will add it later. If you already do, just follow the working.

$\int {\sec}^{3} u = \frac{1}{2} \sec u \tan u + \frac{1}{2} \ln | \sec u + \tan u | + \text{C}$

$\int {\sec}^{3} u$ $\mathrm{du} - \int \sec u$ $\mathrm{du} =$
$\frac{1}{2} \sec u \tan u + \frac{1}{2} \ln | \sec u + \tan u | + \text{C"-ln|secu+tanu|+"C}$
$= \frac{1}{2} \sec u \tan u - \frac{1}{2} \ln | \sec u + \tan u | + \text{c}$

Now all that's left is to rewrite the answer in terms of $x$.

$\frac{1}{2} \sec u \tan u - \frac{1}{2} \ln | \sec u + \tan u | + \text{c} =$

$\frac{1}{2} x \sqrt{{x}^{2} - 1} - \frac{1}{2} \ln | x + \sqrt{{x}^{2} - 1} | + \text{c}$

How to evaluate $\int {\sec}^{3} u$ $\mathrm{du}$:

$\int {\sec}^{3} u$ $\mathrm{du} = \int {\sec}^{2} u \sec u$ $\mathrm{du}$

$f \left(x\right) = \sec u \Rightarrow f ' \left(x\right) = \sec u \tan u$
$g ' \left(x\right) = {\sec}^{2} u \Rightarrow g \left(x\right) = \tan u$

$\int {\sec}^{2} u \sec u$ $\mathrm{du} = \sec u \tan u - \int \sec u {\tan}^{2} u$ $\mathrm{du} =$
$\sec u \tan u - \int \sec u \left({\sec}^{2} u - 1\right)$ $\mathrm{du} =$
secutanu-(intsec^3u $\mathrm{du} - \int \sec u$ du)=

$\int {\sec}^{3} u$ $\mathrm{du} = \sec u \tan u - \int {\sec}^{3} u$ $\mathrm{du} + \int \sec u$ $\mathrm{du}$

$2 \int {\sec}^{3} u$ $\mathrm{du} = \sec u \tan u + \ln | \sec u + \tan u |$

$\int {\sec}^{3} u$ $\mathrm{du} = \frac{1}{2} \left(\sec u \tan u + \ln | \sec u + \tan u |\right) + \text{C}$