# How do you integrate int e^(2x)/sqrt(-e^(2x) -25)dx using trigonometric substitution?

Apr 9, 2018

color(red)(-sqrt(-e^(2x)-25)+C

#### Explanation:

First factor out $i$ from the root.

$\int {e}^{2 x} / \sqrt{- {e}^{2 x} - 25} \mathrm{dx}$

$\int {e}^{2 x} / \sqrt{- 1 \left({e}^{2 x} + 25\right)} \mathrm{dx}$

$\int {e}^{2 x} / \left(\sqrt{- 1} \sqrt{{e}^{2 x} + 25}\right) \mathrm{dx}$

$\frac{1}{i} \int {e}^{2 x} / \sqrt{{e}^{2 x} + 25} \mathrm{dx}$

$- i \int {e}^{2 x} / \sqrt{{e}^{2 x} + 25} \mathrm{dx}$

Now we can perform trig substitution.

It's important to recognize:

${e}^{2 x} = {e}^{x} \cdot {e}^{x}$

Let

$\tan \theta = {e}^{x} / 5$

${e}^{x} = 5 \tan \theta$

${e}^{x} \cdot \mathrm{dx} = 5 {\sec}^{2} \theta \cdot d \theta$

$\sec \theta = \frac{\sqrt{{e}^{2 x} + 25}}{5}$

$5 \sec \theta = \sqrt{{e}^{2 x} + 25}$

Substituting, we get:

$- i \int \frac{5 \tan \theta}{5 \sec \theta} 5 {\sec}^{2} \theta \cdot d \theta$

$- i \int 5 \tan \theta \sec \theta \cdot d \theta$

Integrating gives us:

$- i \left[5 \sec \theta\right] + C$

$- i \left[\sqrt{{e}^{2 x} + 25}\right] + C$

color(red)(-sqrt(-e^(2x)-25)+C