# How do you integrate int 1/sqrt(-e^(2x)-20e^x-136)dx using trigonometric substitution?

Jun 6, 2018

$\int \setminus \frac{1}{\sqrt{- {e}^{2 x} - 20 {e}^{x} - 136}} \setminus \mathrm{dx} = - \frac{i}{2 \sqrt{34}} \setminus \text{arcsinh} \setminus \left(\frac{68}{3} {e}^{- x} + \frac{5}{3}\right) + C$

#### Explanation:

We seek:

$I = \frac{1}{\sqrt{- {e}^{2 x} - 20 {e}^{x} - 136}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{e}^{2 x} \left(- 1 - 20 {e}^{- x} - 136 {e}^{- 2 x}\right)}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{{e}^{x} \sqrt{\left(- 1 - 20 {e}^{- x} - 136 {e}^{- 2 x}\right)}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{{e}^{- x}}{\sqrt{- 1 - 20 {e}^{- x} - 136 {e}^{- 2 x}}} \setminus \mathrm{dx}$

We can perform a substitution, Let:

$u = - 136 {e}^{- x} - 10 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 136 {e}^{- x}$, and, ${e}^{- x} = - \frac{u + 10}{136}$

The we can write the integral as:

$I = \int \setminus \frac{\left(\frac{1}{136}\right)}{\sqrt{- 1 + 20 \left(\frac{u + 10}{136}\right) - 136 {\left(\frac{u + 10}{136}\right)}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{136} \setminus \int \setminus \frac{1}{\sqrt{- 1 + \frac{20}{136} \left(u + 10\right) - \frac{1}{136} {\left(u + 10\right)}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{136} \setminus \int \setminus \frac{1}{\sqrt{\frac{1}{136} \left(- 136 + 20 \left(u + 10\right) - {\left(u + 10\right)}^{2}\right)}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{\sqrt{136}}{136} \setminus \int \setminus \frac{1}{\sqrt{- 136 + 20 u + 200 - {u}^{2} - 20 u - 100}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{\sqrt{136}} \setminus \int \setminus \frac{1}{\sqrt{- {u}^{2} - 36}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{2 \sqrt{34}} \setminus \int \setminus \frac{1}{\sqrt{\left(- 1\right) \left({u}^{2} + {6}^{2}\right)}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{2 \sqrt{34}} \setminus \int \setminus \frac{1}{i \sqrt{\left({u}^{2} + {6}^{2}\right)}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{2 \sqrt{34}} \setminus \int \setminus \frac{- i}{\sqrt{\left({u}^{2} + {6}^{2}\right)}} \setminus \mathrm{du}$

This is a standard integral, so we can write (temporarily omitting the constant of integration):

$I = - \frac{i}{2 \sqrt{34}} \setminus \text{arcsinh} \setminus \left(\frac{u}{6}\right)$

And if we restore the substitution, we get:

$I = - \frac{i}{2 \sqrt{34}} \setminus \text{arcsinh} \setminus \left(\frac{136 {e}^{- x} + 10}{6}\right) + C$

$\setminus \setminus = - \frac{i}{2 \sqrt{34}} \setminus \text{arcsinh} \setminus \left(\frac{68}{3} {e}^{- x} + \frac{5}{3}\right) + C$