How do you integrate int 1/sqrt(e^(2x)+12e^x+40)dx using trigonometric substitution?

May 23, 2018

$\int \setminus \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 40}} \setminus \mathrm{dx} = - \frac{\sqrt{10}}{20} \setminus \text{arcsinh} \setminus \left(20 {e}^{- x} + 3\right) + C$

Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 40}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{e}^{2 x} \left(1 + 12 {e}^{- x} + 40 {e}^{- 2 x}\right)}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{{e}^{x} \sqrt{1 + 12 {e}^{- x} + 40 {e}^{- 2 x}}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus {e}^{- x} / \left(\sqrt{1 + 12 {e}^{- x} + 40 {e}^{- 2 x}}\right) \setminus \mathrm{dx}$

We can perform a substitution, Let:

$u = 40 {e}^{- x} + 6 \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 40 {e}^{- x}$, and, ${e}^{- x} = \frac{u - 6}{40}$

The we can write the integral as:

$I = \int \setminus \frac{- \frac{1}{40}}{\sqrt{1 + 12 \left(\frac{u - 6}{40}\right) + 40 {\left(\frac{u - 6}{40}\right)}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{40} \setminus \int \setminus \frac{1}{\sqrt{1 + 12 \left(\frac{u - 6}{40}\right) + {\left(u - 6\right)}^{2} / 40}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{40} \setminus \int \setminus \frac{1}{\sqrt{\frac{1}{40} \left(40 + 12 \left(u - 6\right) + {\left(u - 6\right)}^{2}\right)}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{40} \setminus \int \setminus \frac{1}{\sqrt{\frac{1}{40}} \sqrt{40 + 12 u - 72 + {u}^{2} - 12 u + 36}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{\sqrt{40}}{40} \setminus \int \setminus \frac{1}{\sqrt{4 + {u}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{\sqrt{40}}{40} \setminus \int \setminus \frac{1}{\sqrt{{2}^{2} + {u}^{2}}} \setminus \mathrm{du}$

This is a standard integral, so we can write:

$I = - \frac{\sqrt{40}}{40} \setminus \text{arcsinh} \setminus \left(\frac{u}{2}\right) + C$

And if we restore the substitution, we get:

$I = - \frac{\sqrt{40}}{40} \setminus \text{arcsinh} \setminus \left(\frac{40 {e}^{- x} + 6}{2}\right) + C$

$\setminus \setminus = - \frac{\sqrt{10}}{20} \setminus \text{arcsinh} \setminus \left(20 {e}^{- x} + 3\right) + C$