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

Nov 29, 2016

$\ln \left({e}^{x} + \sqrt{{e}^{2 x} - 1}\right) - {\sec}^{-} 1 \left({e}^{x}\right) + C$

#### Explanation:

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

Apply the substitution ${e}^{x} = \sec \theta$. This implies that ${e}^{x} \mathrm{dx} = \sec \theta \tan \theta d \theta$. First rearranging, then substituting, we see that:

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

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

Using ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$I = \int \left(\sec \theta - 1\right) d \theta$

$I = \ln \left\mid \sec \theta + \tan \theta \right\mid - \theta$

$I = \ln \left\mid \sec \theta + \sqrt{{\sec}^{2} \theta - 1} \right\mid - \theta$

Using ${e}^{x} = \sec \theta$ and $\theta = {\sec}^{-} 1 \left({e}^{x}\right)$:

$I = \ln \left\mid {e}^{x} + \sqrt{{e}^{2 x} - 1} \right\mid - {\sec}^{-} 1 \left({e}^{x}\right)$

Since ${e}^{x} + \sqrt{{e}^{2 x} - 1} > 0$ for all values of $x$, the absolute value bars aren't needed:

$I = \ln \left({e}^{x} + \sqrt{{e}^{2 x} - 1}\right) - {\sec}^{-} 1 \left({e}^{x}\right) + C$