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

Jul 22, 2015

With this general case, notice how $\sqrt{{x}^{2} - {a}^{2}} \propto \sqrt{{\sec}^{2} \theta - 1}$. So, we can use the following substitution:

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

Thus, we have:

$= \int \frac{1}{\cancel{a \tan \theta}} \cdot \cancel{a} \sec \theta \cancel{\tan \theta} d \theta$

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

Then just a little trick:
$= \int \sec \theta \left(\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\right) d \theta$

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

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

Therefore:

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

$= \ln | u |$

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

Recall that:
$\sec \theta = \frac{x}{a}$
$\tan \theta = \frac{\sqrt{{x}^{2} - {a}^{2}}}{a}$

Thus we have:

$= \textcolor{g r e e n}{\ln | \frac{x}{a} + \frac{\sqrt{{x}^{2} - {a}^{2}}}{a} | + C}$

...which is perfectly acceptable. But, you could simplify this more and be a little sneaky.

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

$= \ln | x + \sqrt{{x}^{2} - {a}^{2}} | + \ln | \frac{1}{a} | + C$

but since $a$ is a constant... it gets embedded into $C$.

$= \textcolor{b l u e}{\ln | x + \sqrt{{x}^{2} - {a}^{2}} | + C}$

So if you see Wolfram Alpha give you this answer, that's why.