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

Apr 19, 2018

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \text{arcosh"(x/a)+"c}$

#### Explanation:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \int \frac{\mathrm{dx}}{a \sqrt{{x}^{2} / {a}^{2} - 1}} = \int \frac{1}{a \sqrt{{\left(\frac{x}{a}\right)}^{2} - 1}} \mathrm{dx}$

Now, let $u = \frac{x}{a}$ and $\mathrm{du} = \frac{1}{a} \mathrm{dx}$

Then

$\int \frac{1}{a \sqrt{{\left(\frac{x}{a}\right)}^{2} - 1}} \mathrm{dx} = \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{dx} = \text{arcosh"(u)+"c"="arcosh"(x/a)+"c}$

Apr 19, 2018

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

#### Explanation:

The integrand is defined for $x \in \left(- \infty , - a\right) \cup \left(a , + \infty\right)$. Let us focus first on $x \in \left(a , + \infty\right)$ and substitute:

$x = a \sec t$

$\mathrm{dx} = a \sec t \tan t$

with $t \in \left(0 , \frac{\pi}{2}\right)$

so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \int \frac{a \sec t \tan t \mathrm{dt}}{\sqrt{{a}^{2} {\sec}^{2} t - {a}^{2}}}$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Use now the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

and as for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive:

sqrt(sec^2t-1) = tan^

Then:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \int \sec t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \ln \left\mid \sec t + \tan t \right\mid + C$

Undoing the substitution:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}} = \ln \left\mid \frac{x}{a} + \sqrt{{x}^{2} / {a}^{2} - 1} \right\mid + C$

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

By differentiating we can see that the solution is valid also for $x \in \left(- \infty , - 3\right)$