# How do you integrate int 1/sqrt(x^2-9) by trigonometric substitution?

Dec 11, 2016

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 9}} = \log | \frac{x}{3} + \frac{1}{3} \sqrt{{x}^{9} - 9} |$

#### Explanation:

Substitute:

$x = 3 \sec t$
$\mathrm{dx} = 3 \sec t \tan t \mathrm{dt}$

and you have for $t \in \left(0 , \frac{\pi}{2}\right)$:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 9}} = \int \frac{3 \sec t \tan t \mathrm{dt}}{\sqrt{9 {\sec}^{2} t - 9}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{\frac{1}{\cos} ^ 2 t - 1}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{\frac{1 - {\cos}^{2} t}{\cos} ^ 2 t}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\sin}^{2} \frac{t}{\cos} ^ 2 t}} = \int \frac{\sec t \tan t \mathrm{dt}}{\sqrt{{\tan}^{2} t}}$

In this interval for $\tan t > 0$, so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 9}} = \int \frac{\sec t \tan t \mathrm{dt}}{\tan} t = \int \sec t \mathrm{dt} = \log | \sec t + \tan t |$

To go back to x we note that:

$\sec t = \frac{x}{3}$

and $\tan t = \sin t \cdot \sec t = \sec t \sqrt{1 - \frac{1}{\sec} ^ t 2 t} = \frac{x}{3} \sqrt{1 - \frac{9}{x} ^ 2} = \frac{1}{3} \sqrt{{x}^{9} - 9}$.

Finally:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 9}} = \log | \frac{x}{3} + \frac{1}{3} \sqrt{{x}^{9} - 9} |$

Dec 11, 2016

Given: $\int \frac{1}{\sqrt{{x}^{2} - 9}} \mathrm{dx}$

Let $x = 3 \sec \left(u\right) \text{, then } \mathrm{dx} = 3 \tan \left(u\right) \sec \left(u\right) \mathrm{du}$

Substitute this into the integral:

$\int \frac{3 \tan \left(u\right) \sec \left(u\right)}{\sqrt{{\left(3 \sec \left(u\right)\right)}^{2} - 9}} \mathrm{du} =$

$\int \frac{3 \tan \left(u\right) \sec \left(u\right)}{\sqrt{9 {\sec}^{2} \left(u\right) - 9}} \mathrm{du} =$

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

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

$\int \frac{\tan \left(u\right) \sec \left(u\right)}{\sqrt{{\tan}^{2} \left(u\right)}} \mathrm{du} =$

$\int \frac{\tan \left(u\right) \sec \left(u\right)}{\tan} \left(u\right) \mathrm{du} =$

This integral is in any good list of integrals:

$\int \sec \left(u\right) \mathrm{du} = \ln \left(\tan \left(u\right) + \sec \left(u\right)\right) + C$

$\int \frac{1}{\sqrt{{x}^{2} - 9}} \mathrm{dx} = \ln \left(\sqrt{{\sec}^{2} \left(u\right) - 1} + \sec \left(u\right)\right) + C$

Substitute $\frac{x}{3}$ for $\sec \left(u\right)$:

$\int \frac{1}{\sqrt{{x}^{2} - 9}} \mathrm{dx} = \ln \left(\sqrt{{x}^{2} / 9 - 1} + \frac{x}{3}\right) + C$

Dec 11, 2016

${\cosh}^{- 1} | \frac{x}{3} | + C$

#### Explanation:

Substitute $x = 3 \cosh u$, $\mathrm{dx} = 3 \sinh u \mathrm{du}$ and use ${\cosh}^{2} u - {\sinh}^{2} u = 1$ to get the integral as being $u$, which is ${\cosh}^{- 1} \left(\frac{x}{3}\right)$
which can be written as $\ln | x + \sqrt{{x}^{2} - 9} | + C$ which is equivalent to
$\ln | \left(\frac{x}{3} + \left(\frac{1}{3}\right) \sqrt{{x}^{2} -} 9\right) |$ given above, as the two answers differ by a constant of $\ln 3$ which is subsumed into the constant of integration.