# How do you find the integral of (x^2 - 1)^(1/2)?

Jun 30, 2015

First of all, notice how this equation has this relationship:

$\sqrt{{x}^{2} - 1} \propto \sqrt{{a}^{2} {x}^{2} - {a}^{2}} \propto \sqrt{{a}^{2} {\sec}^{2} \theta - {a}^{2}}$
where $\propto$ means "proportional to", and $a = 1$. Thus, let:

$a = 1$
$x = a \sec \theta = \sec \theta$
$\mathrm{dx} = \sec \theta \tan \theta d \theta$
$\sqrt{{x}^{2} - 1} = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{\tan}^{2} \theta} = \tan \theta$

Now you can write this as:

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

$= \int \tan \theta \sec \theta \tan \theta d \theta$

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

Something you can do to simplify this (and it may seem odd at first) is to rewrite this as:

$= \int \sec \theta \left({\sec}^{2} \theta - 1\right) d \theta$

since ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$. Now we only have $\sec \theta$ to deal with.

$= \int {\sec}^{3} \theta - \sec \theta d \theta$

These two integrals require special tricks, but they do have real answers. Let's take them separately for simplicity.

$\int \sec \theta d \theta$

$= \int \left(\sec \theta\right) \left(\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\right) d \theta$

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

Now notice that if you take the derivative of the denominator, you get the numerator times $d \theta$. i.e.:

$d \left[\sec \theta + \tan \theta\right] = \left({\sec}^{2} \theta + \sec \theta \tan \theta\right) d \theta$

Thus, let:
$u = \sec \theta + \tan \theta$
$\mathrm{du} = \sec \theta \tan \theta + {\sec}^{2} \theta d \theta$

$\implies \int \frac{1}{u} \mathrm{du}$

$= \ln | u | = \ln | \sec \theta + \tan \theta |$

Now let's take the other integral.

$\int {\sec}^{3} \theta d \theta$

With this, the best trick one can try is Integration by Parts. Typically, you let $u = \sec \theta$, so let:

$u = \sec \theta$
$\mathrm{du} = \sec \theta \tan \theta d \theta$
$\mathrm{dv} = {\sec}^{2} \theta d \theta$
$v = \tan \theta$

$= \sec \theta \tan \theta - \int \sec \theta {\tan}^{2} \theta d \theta$

Look at that, we got back the original integral (didn't even need an identity for this step)!

The whole thing now is:

$\int \sec \theta {\tan}^{2} \theta d \theta$

$= \sec \theta \tan \theta - \textcolor{\mathrm{da} r k g r e e n}{\int \sec \theta {\tan}^{2} \theta d \theta} - \int \sec \theta d \theta$

$2 \int \sec \theta {\tan}^{2} \theta d \theta = \sec \theta \tan \theta - \int \sec \theta d \theta$

Technically, we're done. Now we just have:

$\int \sec \theta {\tan}^{2} \theta d \theta = \frac{1}{2} \left[\sec \theta \tan \theta - \int \sec \theta d \theta\right]$

$= \frac{1}{2} \left[\sec \theta \tan \theta - \ln | \sec \theta + \tan \theta |\right]$

Now if you look all the way at the top:
$x = \sec \theta$
$\sqrt{{x}^{2} - 1} = \tan \theta$

Thus:

= color(blue)(1/2[xsqrt(x^2 - 1) - ln|x + sqrt(x^2 - 1)|] + C

You can also see it here.