# How do you integrate int 1/sqrt(e^(2x)+12e^x+37)dx using trigonometric substitution?

Mar 11, 2018

$\frac{\sqrt{37}}{37} L n \left(\frac{\tan \left[\frac{1}{2} \cdot \arctan \left(\frac{{e}^{x} + 1}{6}\right)\right] + 6 - \sqrt{37}}{\tan \left[\frac{1}{2} \cdot \arctan \left(\frac{{e}^{x} + 1}{6}\right)\right] + 6 + \sqrt{37}}\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 37}}$

=$\int \frac{\mathrm{dx}}{\sqrt{{\left({e}^{x} + 1\right)}^{2} + {6}^{2}}}$

=$\int \frac{{e}^{x} \cdot \mathrm{dx}}{{e}^{x} \cdot \sqrt{{\left({e}^{x} + 1\right)}^{2} + {6}^{2}}}$

After using ${e}^{x} + 1 = 6 \tan y$, ${e}^{x} = 6 \tan y - 1$ and ${e}^{x} \cdot \mathrm{dx} = 6 {\left(\sec u\right)}^{2} \cdot \mathrm{du}$ transforms, this integral became

$\int \frac{6 {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{\left(6 \tan y - 1\right) \cdot 6 \sec y}$

=int (secy*dy)/[(6tany-1)

After using $z = \tan \left(\frac{y}{2}\right)$, $\mathrm{dy} = \frac{2 \mathrm{dz}}{{z}^{2} + 1}$, $\tan y = \frac{2 z}{1 - {z}^{2}}$ and $\sec y = \frac{1 + {z}^{2}}{1 - {z}^{2}}$ transforms, it became

=$\int \frac{\frac{1 + {z}^{2}}{1 - {z}^{2}} \frac{2 \mathrm{dz}}{{z}^{2} + 1}}{6 \cdot \frac{2 z}{1 - {z}^{2}} - 1}$

=$\int \frac{\frac{2 z \cdot \mathrm{dz}}{1 - {z}^{2}}}{\frac{{z}^{2} - 12 z - 1}{1 - {z}^{2}}}$

=$\int \frac{2 \mathrm{dz}}{{z}^{2} + 12 z - 1}$

=$\int \frac{2 \mathrm{dz}}{{\left(z + 6\right)}^{2} - 37}$

=$\int \frac{2 \mathrm{dz}}{\left(z + 6 + \sqrt{37}\right) \cdot \left(z + 6 - \sqrt{37}\right)}$

=$\frac{\sqrt{37}}{37} \int \frac{\mathrm{dz}}{z + 6 - \sqrt{37}} - \frac{\sqrt{37}}{37} \int \frac{\mathrm{dz}}{z + 6 + \sqrt{37}}$

=$\frac{\sqrt{37}}{37} L n \left(z + 6 - \sqrt{37}\right)$-$\frac{\sqrt{37}}{37} L n \left(z + 6 + \sqrt{37}\right) + C$

=$\frac{\sqrt{37}}{37} L n \left(\frac{z + 6 - \sqrt{37}}{z + 6 + \sqrt{37}}\right) + C$

=$\frac{\sqrt{37}}{37} L n \left(\frac{\tan \left(\frac{y}{2}\right) + 6 - \sqrt{37}}{\tan \left(\frac{y}{2}\right) + 6 + \sqrt{37}}\right) + C$

After using ${e}^{x} + 1 = 6 \tan y$ and $y = \arctan \left(\frac{{e}^{x} + 1}{6}\right)$ inverse transforms, I found

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 37}}$

=$\frac{\sqrt{37}}{37} L n \left(\frac{\tan \left[\frac{1}{2} \cdot \arctan \left(\frac{{e}^{x} + 1}{6}\right)\right] + 6 - \sqrt{37}}{\tan \left[\frac{1}{2} \cdot \arctan \left(\frac{{e}^{x} + 1}{6}\right)\right] + 6 + \sqrt{37}}\right) + C$