# How do you integrate dx/ sqrt(x^2 - a^2)?

Aug 23, 2015

I takes a couple of substitutions. See explanation.

#### Explanation:

Use a trigonometric substitution: $x = a \sec \theta$

so $\mathrm{dx} = a \sec \theta \tan \theta d \theta$

With a bit of work you can simplify $\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}}$ to

$\int \sec \theta \text{ } d \theta$

If you know this integral, you can skip the next section.

To get that integral multiply by $1$ in the form
$\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$

This gets us:

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

And the numerator is the derivative of the denominator so we get an $\ln$

We get
$\int \sec \theta d \theta = \ln \left(\sec \theta + \tan \theta\right) + C$

Now that we have integrated the secant, note that due to the first substitution, $\sec \theta = \frac{x}{a}$.

Our trigonometry then gets us
$\tan \theta = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{x}^{2} / {a}^{2} - 1}$

$\ln \left(\sec \theta + \tan \theta\right) + C = \ln \left(\frac{x}{a} + \sqrt{{x}^{2} / {a}^{2} - 1}\right) + C$

We can rewrite in several ways. Perhaps the simplest is to write:

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

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

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

But assuming that $a$ is a constant, $\ln a$ is a constant so we can let it be absorbed by the $C$.

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

Checking the answer by differentiating is left as an exercise. (It's not very tedious.)