# How do you integrate int 1/sqrt(-e^(2x)+12e^x-27)dx using trigonometric substitution?

\color{blue}{2/{3\sqrt3}\tan^{-1}(\frac{2\tan (1/2sin^{-1}(\frac{e^x-6}{3}))+1}{\sqrt3})+C

#### Explanation:

$\setminus \int \setminus \frac{1}{\setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}} \setminus \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{9 - \left({e}^{2 x} - 12 {e}^{x} + 36\right)}} \setminus \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{9 - {\left({e}^{x} - 6\right)}^{2}}} \setminus \mathrm{dx}$

Let ${e}^{x} - 6 = 3 \setminus \sin \setminus \theta \setminus \implies {e}^{x} \setminus \mathrm{dx} = 3 \setminus \cos \setminus \theta \setminus d \setminus \theta$

or $\mathrm{dx} = \setminus \frac{\setminus \cos \setminus \theta \setminus d \setminus \theta}{\setminus \sin \setminus \theta + 2}$,

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{9 - 9 \setminus {\sin}^{2} \setminus \theta}} \setminus \setminus \frac{\setminus \cos \setminus \theta \setminus d \setminus \theta}{\setminus \sin \setminus \theta + 2}$

$= \setminus \int \setminus \frac{1}{3 \setminus \cos \setminus \theta} \setminus \setminus \frac{\setminus \cos \setminus \theta \setminus d \setminus \theta}{\setminus \sin \setminus \theta + 2}$

$= \frac{1}{3} \setminus \int \setminus \frac{d \setminus \theta}{\setminus \sin \setminus \theta + 2}$

$= \frac{1}{3} \setminus \int \setminus \frac{d \setminus \theta}{\setminus \frac{2 \setminus \tan \left(\setminus \frac{\theta}{2}\right)}{1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)} + 2}$

$= \frac{1}{6} \setminus \int \setminus \frac{\left(1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)\right) \setminus \setminus d \setminus \theta}{\setminus \tan \left(\setminus \frac{\theta}{2}\right) + 1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)}$

$= \frac{1}{3} \setminus \int \setminus \frac{\frac{1}{2} \setminus {\sec}^{2} \left(\setminus \frac{\theta}{2}\right) \setminus \setminus d \setminus \theta}{{\left(\setminus \tan \left(\setminus \frac{\theta}{2}\right) + \frac{1}{2}\right)}^{2} + \frac{3}{4}}$

$= \frac{1}{3} \setminus \int \setminus \frac{d \left(\setminus \tan \left(\setminus \frac{\theta}{2}\right) + \frac{1}{2}\right)}{{\left(\setminus \tan \left(\setminus \frac{\theta}{2}\right) + \frac{1}{2}\right)}^{2} + {\left(\setminus \frac{\sqrt{3}}{2}\right)}^{2}}$

Using standard formula: $\setminus \int \setminus \frac{\mathrm{dt}}{{t}^{2} + {a}^{2}} = \frac{1}{a} \setminus {\tan}^{- 1} \left(\frac{t}{a}\right)$,

$= \frac{1}{3} \left(\frac{1}{\setminus \frac{\sqrt{3}}{2}}\right) \setminus {\tan}^{- 1} \left(\setminus \frac{\setminus \tan \left(\setminus \frac{\theta}{2}\right) + \frac{1}{2}}{\setminus \frac{\sqrt{3}}{2}}\right) + C$

$= \frac{2}{3 \setminus \sqrt{3}} \setminus {\tan}^{- 1} \left(\setminus \frac{2 \setminus \tan \left(\setminus \frac{\theta}{2}\right) + 1}{\setminus \sqrt{3}}\right) + C$

$= \frac{2}{3 \setminus \sqrt{3}} \setminus {\tan}^{- 1} \left(\setminus \frac{2 \setminus \tan \left(\frac{1}{2} {\sin}^{- 1} \left(\setminus \frac{{e}^{x} - 6}{3}\right)\right) + 1}{\setminus \sqrt{3}}\right) + C$

Jul 27, 2018

$I = \frac{2}{3 \sqrt{3}} a r c \tan \left(\frac{{e}^{x} - 2 \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}}{\sqrt{3} \left({e}^{x} - 6\right)}\right) + c$

#### Explanation:

Here ,

$I = \int \frac{1}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}} \mathrm{dx}$

$- {e}^{2 x} + 12 {e}^{x} - 27 = 9 - {e}^{2 x} + 12 {e}^{x} - 36 = 9 - {\left({e}^{x} - 6\right)}^{2}$

$\therefore I = \int \frac{1}{\sqrt{9 - {\left({e}^{x} - 6\right)}^{2}}} \mathrm{dx}$

Subst. $\textcolor{red}{{e}^{x} - 6 = 3 \sin u} \implies {e}^{x} = 6 + 3 \sin u \implies {e}^{x} \mathrm{dx} = 3 \cos u \mathrm{du}$

=>dx=(3cosudu)/(6+3sinu) andcolor(red)( sinu=(e^x-6)/3

So,

$I = \int \frac{1}{\sqrt{9 - 9 {\sin}^{2} u}} \cdot \frac{3 \cos u \mathrm{du}}{6 + 3 \sin u}$

$= \int \frac{1}{3 \cos u} \times \frac{3 \cos u \mathrm{du}}{6 + 3 \sin u}$

$= \int \frac{1}{6 + 3 \sin u} \mathrm{du} = \frac{1}{3} \int \frac{1}{2 + \sin u} \mathrm{du}$

Subst. color(blue)(tan(u/2)=t=>sec^2(u/2)*1/2du=dt=>du=(2dt)/(1+tan^2(u/2))

$\implies \mathrm{du} = \frac{2 \mathrm{dt}}{1 + {t}^{2}} \mathmr{and} \sin u = \frac{2 t}{1 + {t}^{2}}$

$\therefore I = \frac{1}{3} \int \frac{1}{2 + \frac{2 t}{1 + {t}^{2}}} \times \frac{2 \mathrm{dt}}{1 + {t}^{2}}$

$= \frac{2}{3} \int \frac{1}{2 + 2 {t}^{2} + 2 t} \mathrm{dt} = \frac{1}{3} \int \frac{1}{{t}^{2} + t + 1} \mathrm{dt}$

$= \frac{1}{3} \int \frac{1}{{t}^{2} + t + \frac{1}{4} + \frac{3}{4}} \mathrm{dt} = \frac{1}{3} \int \frac{1}{{\left(t + \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dt}$

$= \frac{1}{3} \frac{1}{\frac{\sqrt{3}}{2}} a r c \tan \left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + c$

$\therefore I = \frac{2}{3 \sqrt{3}} \arctan \left(\frac{2 t + 1}{\sqrt{3}}\right) + c \ldots \ldots \to \left(A\right)$

Now, color(blue)(t=tan(u/2)

=>t^2=tan^2(u/2)=(1-cosu)/(1+cosu)=(1-cosu)/(1+cosu) xxcolor(green)((1-cosu)/(1-cosu))

$\implies {t}^{2} = {\left(1 - \cos u\right)}^{2} / \left(1 - {\cos}^{2} u\right) = {\left(1 - \cos u\right)}^{2} / {\sin}^{2} u$

$\implies t = \frac{1 - \cos u}{\sin} u = \frac{1 - \sqrt{1 - {\sin}^{2} u}}{\sin} u$

Subst. back color(red)(sinu=(e^x-6)/3

$\therefore t = \frac{1 - \sqrt{1 - {\left(\frac{{e}^{x} - 6}{3}\right)}^{2}}}{\frac{{e}^{x} - 6}{3}}$

$\therefore t = \frac{1 - \frac{\sqrt{9 - {\left({e}^{x} - 6\right)}^{2}}}{3}}{\frac{{e}^{x} - 6}{3}} = \frac{3 - \sqrt{- {e}^{2 x} + 12 x - 27}}{{e}^{x} - 6}$

$\therefore 2 t + 1 = \frac{6 - 2 \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}}{{e}^{x} - 6} + 1$

$\therefore 2 t + 1 = \frac{\left(6 - 2 \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27} + {e}^{x} - 6\right)}{{e}^{x} - 6}$

$\therefore 2 t + 1 = \frac{{e}^{x} - 2 \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}}{{e}^{x} - 6}$

So, from $\left(A\right)$

$I = \frac{2}{3 \sqrt{3}} a r c \tan \left(\frac{{e}^{x} - 2 \sqrt{- {e}^{2 x} + 12 {e}^{x} - 27}}{\sqrt{3} \left({e}^{x} - 6\right)}\right) + c$