# How do you integrate int 1/sqrt(e^(2x)-2e^x)dx using trigonometric substitution?

Feb 1, 2017

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} - 2 {e}^{x}}} = \sqrt{1 - 2 {e}^{- x}} + C$

#### Explanation:

Complete the square under the root:

$\frac{1}{\sqrt{{e}^{2 x} - 2 {e}^{x}}} = \frac{1}{\sqrt{{e}^{2 x} - 2 {e}^{x} + 1 - 1}} = \frac{1}{\sqrt{{\left({e}^{x} - 1\right)}^{2} - 1}}$

Now substitute:

${e}^{x} - 1 = t$

$x = \ln \left(1 + t\right)$

$\mathrm{dx} = \frac{\mathrm{dt}}{1 + t}$

We have:

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} - 2 {e}^{x}}} = \int \frac{\mathrm{dx}}{\sqrt{{\left({e}^{x} - 1\right)}^{2} - 1}} = \int \frac{\mathrm{dt}}{\left(1 + t\right) \sqrt{{t}^{2} - 1}}$

Substitute again:

$t = \sec u$

$\mathrm{dt} = \sec u \tan u \mathrm{du}$

to have:

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

Using the identity:

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

this becomes:

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

Use now:

${\cos}^{2} \left(\frac{u}{2}\right) = \frac{1 + \cos u}{2}$

$\int \frac{\mathrm{du}}{1 + \cos u} = \frac{1}{2} \int \frac{\mathrm{du}}{\cos} ^ 2 \left(\frac{u}{2}\right) = \int \frac{d \left(\frac{u}{2}\right)}{\cos} ^ 2 \left(\frac{u}{2}\right) = \tan \left(\frac{u}{2}\right) + C$

To undo the substitutions, consider first the identity:

$\tan \left(\frac{u}{2}\right) = \sqrt{\frac{1 - \cos u}{1 + \cos u}}$

and as $\cos u = \frac{1}{\sec} u = \frac{1}{t}$:

$\tan \left(\frac{u}{2}\right) = \sqrt{\frac{1 - \frac{1}{t}}{1 + \frac{1}{t}}} = \sqrt{\frac{t - 1}{t + 1}} = \sqrt{\frac{{e}^{x} - 2}{e} ^ x} = \sqrt{1 - 2 {e}^{- x}}$

so that eventually:

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} - 2 {e}^{x}}} = \sqrt{1 - 2 {e}^{- x}} + C$