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

May 29, 2018

$\int \setminus \frac{1}{\sqrt{{e}^{2 x} - 2 {e}^{x} + 2}} \setminus \mathrm{dx} = - \frac{\sqrt{2}}{2} \setminus \text{arcsinh} \setminus \left(2 {e}^{- x} - 1\right) + C$

#### Explanation:

We seek:

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

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

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

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

We can perform a substitution, Let:

$u = 2 {e}^{- x} - 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 {e}^{- x}$, and, ${e}^{- x} = \frac{u + 1}{2}$

The we can write the integral as:

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

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

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

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

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

This is a standard integral, so we can write:

$I = - \frac{\sqrt{2}}{2} \setminus \text{arcsinh} \setminus \left(u\right) + C$

And if we restore the substitution, we get:

$I = - \frac{\sqrt{2}}{2} \setminus \text{arcsinh} \setminus \left(2 {e}^{- x} - 1\right) + C$