# How do you integrate int 1/sqrt(e^(2x)-2e^x+10)dx using trigonometric substitution?

Feb 24, 2016

$\int \frac{1}{\sqrt{{e}^{2 x} - 2 {e}^{x} + 10}} \mathrm{dx}$

$\int \frac{{e}^{x} {e}^{-} x}{\sqrt{{e}^{2 x} - 2 {e}^{x} + 10}} \mathrm{dx}$

$t = {e}^{x}$
$\mathrm{dt} = {e}^{x}$

$\int \frac{1}{t \sqrt{{t}^{2} - 2 t + 10}} \mathrm{dt}$

$\int \frac{1}{t \sqrt{{t}^{2} - 2 t + 1 + 9}} \mathrm{dt}$

$\int \frac{1}{t \sqrt{{\left(t - 1\right)}^{2} + 9}} \mathrm{dt}$

$t - 1 = 3 \tan \left(u\right)$
$\mathrm{dt} = 3 \left({\tan}^{2} \left(u\right) + 1\right) \mathrm{du}$

${\left(t - 1\right)}^{2} = 9 {\tan}^{2} \left(u\right)$

$\int \frac{1}{t \sqrt{{\left(t - 1\right)}^{2} + 9}} \mathrm{dx}$

$\int \frac{\frac{3}{\cos} ^ 2 \left(u\right)}{\left(3 \tan \left(u\right) + 1\right) \left(\frac{3}{\cos} \left(u\right)\right)} \mathrm{du}$

$\int \frac{\frac{3}{\cos} ^ 2 \left(u\right)}{\left(9 \sin \frac{u}{\cos} ^ 2 \left(u\right) + \frac{3}{\cos} \left(u\right)\right)} \mathrm{du}$

$\int \frac{1}{\left(3 \sin \left(u\right) + \cos \left(u\right)\right)} \mathrm{du}$

$w = \tan \left(\frac{u}{2}\right)$

$3 \sin \left(u\right) = \frac{6 w}{{w}^{2} + 1}$

$\cos \left(u\right) = \frac{1 - {w}^{2}}{{w}^{2} + 1}$

$\mathrm{du} = \frac{2 \mathrm{dw}}{{w}^{2} + 1}$

$\int \frac{\frac{2}{{w}^{2} + 1}}{\left(\frac{6 w}{{w}^{2} + 1} + \frac{1 - {w}^{2}}{{w}^{2} + 1}\right)} \mathrm{dw}$

$\int \frac{2}{6 w + 1 - {w}^{2}} \mathrm{dw}$

$- \int \frac{2}{{w}^{2} - 6 w + 9 - 10} \mathrm{dw}$

$- 2 \int \frac{1}{{\left(w - 3\right)}^{2} - 10} \mathrm{dw}$

$\sqrt{10} v = \left(w - 3\right)$
$\sqrt{10} \mathrm{dv} = \mathrm{dw}$

$- 2 \frac{\sqrt{10}}{10} \int \frac{1}{{v}^{2} - 1} \mathrm{dv}$

$- 2 \frac{\sqrt{10}}{10} \left[\arctan h \left(v\right)\right] + C$

$- 2 \frac{\sqrt{10}}{10} \left[\arctan h \left(\frac{\tan \left(\arctan \frac{\frac{{e}^{x} - 1}{3}}{2}\right) - 3}{\sqrt{10}}\right)\right] + C$