# How do you integrate int e^(x)/sqrt(e^(2x) -81)dx using trigonometric substitution?

Jul 30, 2016

$\int \frac{{e}^{x}}{\sqrt{{e}^{2 x} - 81}} \mathrm{dx} = \ln \left(\frac{1}{9} \left(\sqrt{{e}^{2 x} - 81} + {e}^{x}\right)\right) + C$

#### Explanation:

First, we make the substitution $u = {e}^{x} \implies \mathrm{du} = {e}^{x} \mathrm{dx}$

Integral becomes:

$\int \frac{\mathrm{du}}{\sqrt{{u}^{2} - 81}}$

Now, whatever trig substitution we use, we want the coefficient to be 9. This is because squaring 9 gives 81 which means we can then move a factor of 9 outside the square root.

Let $u = 9 \sec \left(\theta\right) \implies \mathrm{du} = 9 \sec \left(\theta\right) \tan \left(\theta\right) d \theta$

$\int \frac{9 \sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{81 {\sec}^{2} \left(\theta\right) - 81}} d \theta$

$\int \frac{\cancel{9} \sec \left(\theta\right) \tan \left(\theta\right)}{\cancel{9} \sqrt{{\sec}^{2} \left(\theta\right) - 1}} d \theta = \int \frac{\sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{{\sec}^{2} \left(\theta\right) - 1}} d \theta$

We know the first Pythagorean identity:

${\sin}^{2} \phi + {\cos}^{2} \phi = 1$

Dividing every term by ${\cos}^{2} \phi$ gives the second identity:

${\tan}^{2} \phi + 1 = {\sec}^{2} \phi \implies {\tan}^{2} \phi = {\sec}^{2} \phi - 1$

So our integral becomes:

$\int \frac{\sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{{\tan}^{2} \left(\theta\right)}} d \theta = \int \sec \left(\theta\right) d \theta$

This is a standard integral but does require some tricks to do if you don't have a table of integrals. It's only tangentially related to the question at hand though so I'll leave this link which has a very nice method of solving it.

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

$u = 9 \sec \left(\theta\right) \implies \theta = {\sec}^{- 1} \left(\frac{u}{9}\right)$

$= \ln \left(\tan \left({\sec}^{- 1} \left(\frac{u}{9}\right)\right) + \sec \left({\sec}^{- 1} \left(\frac{u}{9}\right)\right)\right) + C$

Now, to tackle $\tan \left({\sec}^{- 1} \left(\frac{u}{9}\right)\right)$. I always find it very helpful to draw a triangle for this kind of thing but it's a bit time consuming so we'll use a quick and dirty method here..

Consider $y = {\sec}^{- 1} \left(x\right) \implies x = \sec \left(y\right)$

$\therefore {x}^{2} = {\sec}^{2} \left(y\right)$

From the earlier pythagorean identity:

${x}^{2} = {\tan}^{2} \left(y\right) + 1$

${x}^{2} - 1 = {\tan}^{2} \left(y\right)$

$\sqrt{{x}^{2} - 1} = \tan \left(y\right) = \tan \left({\sec}^{- 1} \left(x\right)\right)$

$= \ln \left(\sqrt{{\left(\frac{u}{9}\right)}^{2} - 1} + \frac{u}{9}\right) + C$
$= \ln \left(\frac{1}{9} \left(\sqrt{{u}^{2} - 81} + u\right)\right) + C$
$= \ln \left(\frac{1}{9} \left(\sqrt{{e}^{2 x} - 81} + {e}^{x}\right)\right) + C$