# How do you integrate int e^x/sqrt(-e^(2x)-20e^x-96)dx using trigonometric substitution?

Feb 25, 2016

${\sin}^{- 1} \left(\frac{{e}^{x} + 10}{2}\right) + c o n s t .$

#### Explanation:

Considering the expression under the radical sign

$- \left({e}^{x} + 10\right) = - {e}^{2 x} - 20 {e}^{x} - 100$
$\to - {e}^{2 x} - 20 {e}^{x} - 96 = {2}^{2} - \left({e}^{x} + 10\right)$

Now we can see a convenient trigonometric substitution such as:

${e}^{x} + 10 = 2 \sin y$
$\to {e}^{x} \cdot \mathrm{dx} = 2 \cos y \cdot \mathrm{dy}$

$\implies \int {e}^{x} / \sqrt{- {e}^{2 x} - 20 {e}^{x} - 96} \mathrm{dx} = \int \frac{2 \cos y}{\sqrt{4 - {\left(2 \sin y\right)}^{2}}} \mathrm{dy}$
$= \int \frac{\cancel{2} \cos y}{\cancel{2} \sqrt{1 - {\sin}^{2} y}} \mathrm{dy} = \int \frac{\cancel{\cos y}}{\cancel{\cos y}} \mathrm{dy} = \int \mathrm{dy} = y + c o n s t .$

But

$2 \sin y = {e}^{x} + 10$ => $\sin y = \frac{{e}^{x} + 10}{2}$ => $y = {\sin}^{- 1} \left(\frac{{e}^{x} + 10}{2}\right)$

Then the result is
${\sin}^{- 1} \left(\frac{{e}^{x} + 10}{2}\right) + c o n s t .$