# How do you integrate int 1/sqrt(-e^(2x)+12e^x-11)dx using trigonometric substitution?

Mar 26, 2018

$\int \frac{\mathrm{dx}}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 11}} = - \frac{\sqrt{11}}{11} \arcsin \left(\frac{11 {e}^{- x} - 6}{5}\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 11}}$

=$\int \frac{{e}^{- x} \cdot \mathrm{dx}}{\sqrt{- 11 {e}^{- 2 x} + 12 {e}^{- x} - 1}}$

=$\sqrt{11} \int \frac{{e}^{- x} \cdot \mathrm{dx}}{\sqrt{- 121 {e}^{- 2 x} + 132 {e}^{- x} - 11}}$

=$\sqrt{11} \int \frac{{e}^{- x} \cdot \mathrm{dx}}{\sqrt{{5}^{2} - {\left(11 {e}^{- x} - 6\right)}^{2}}}$

=$- \frac{\sqrt{11}}{11} \int \frac{- 11 {e}^{- x} \cdot \mathrm{dx}}{\sqrt{{5}^{2} - {\left(11 {e}^{- x} - 6\right)}^{2}}}$

After using $11 {e}^{- x} - 6 = 5 \sin y$ and $- 11 {e}^{- x} \cdot \mathrm{dx} = 5 \cos y \cdot \mathrm{dy}$ transforms, this integral became

$- \frac{\sqrt{11}}{11} \int \frac{5 \cos y \cdot \mathrm{dy}}{\sqrt{{5}^{2} - {\left(5 \sin y\right)}^{2}}}$

=$- \frac{\sqrt{11}}{11} \int \frac{5 \cos y \cdot \mathrm{dy}}{5 \cos y}$

=$- \frac{\sqrt{11}}{11} \int \mathrm{dy}$

=$- \frac{\sqrt{11}}{11} y + C$

After using $11 {e}^{- x} - 6 = 5 \sin y$ and $y = \arcsin \left(\frac{11 {e}^{- x} - 6}{5}\right)$ inverse transforms, I found

$\int \frac{\mathrm{dx}}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 11}} = - \frac{\sqrt{11}}{11} \arcsin \left(\frac{11 {e}^{- x} - 6}{5}\right) + C$