# How do you integrate int 1/sqrt(-e^(2x)+12e^x+72)dx using trigonometric substitution?

Jul 20, 2018

$I$=$\frac{1}{6 \sqrt{2}} \ln | \frac{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72} + \left(\sqrt{3} - \sqrt{2}\right) \left({e}^{x} - 6\right)}{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72} + \left(\sqrt{3} + \sqrt{2}\right) \left({e}^{x} - 6\right)} |$$+ c$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{- {e}^{2 x} + 12 {e}^{x} + 72}} \mathrm{dx}$

Now,

-e^(2x)+12e^x+72=color(brown)(108)-e^(2x)+12e^x- color(brown)(36)=108-(e^x-6)^2

$\therefore I = \int \frac{1}{\sqrt{108 - {\left({e}^{x} - 6\right)}^{2}}} \mathrm{dx}$

Substitute ,color(red)(e^x-6=sqrt108sinu)=>e^x=sqrt108 sinu+6

$\implies {e}^{x} \mathrm{dx}$=$\sqrt{108} \cos u \mathrm{du} \implies \mathrm{dx} = \frac{\sqrt{108} \sin u}{{e}^{x}}$= (sqrt108cosu)/(sqrt108sinu+6)

So. I=int1/sqrt(108-108sin^2u)* (sqrt108cosu)/(sqrt108sinu+6)du

$I = \int \frac{1}{\sqrt{108} \cos u} \times \frac{\sqrt{108} \cos u}{\sqrt{108} \sin u + 6} \mathrm{du}$

$= \int \frac{1}{\sqrt{108} \sin u + 6} \mathrm{du}$

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

Substitute , $\textcolor{b l u e}{\tan \left(\frac{u}{2}\right) = t} \implies {\sec}^{2} \left(\frac{u}{2}\right) \frac{1}{2} \mathrm{du} = \mathrm{dt}$

$\implies \mathrm{du} = \frac{2 \mathrm{dt}}{1 + {\tan}^{2} \left(\frac{u}{2}\right)} \implies \textcolor{b l u e}{\mathrm{du} = \frac{2 \mathrm{dt}}{1 + {t}^{2}}} \mathmr{and}$

sinu=(2tan(u/2))/(1+tan^2(u/2))=>color(blue)(sinu=(2t)/(1+t^2)

So , $I = \frac{1}{6} \int \frac{1}{\sqrt{3} \left(\frac{2 t}{1 + {t}^{2}}\right) + 1} \cdot \frac{2 \mathrm{dt}}{1 + {t}^{2}}$

$\implies I = \frac{2}{6} \int \frac{1}{2 \sqrt{3} t + 1 + {t}^{2}} \mathrm{dt}$

$\implies I = \frac{2}{6} \int \frac{1}{{t}^{2} + 2 \sqrt{3} t + 3 - 2} \mathrm{dt} \to$[completing square]

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

=>I=1/3*1/(2sqrt2)int [((t+sqrt3+sqrt2)-(t+sqrt3- sqrt2))/((t+sqrt3+sqrt2)(t+sqrt3-sqrt2))]dt

$= \frac{1}{6 \sqrt{2}} \int \left[\frac{1}{t + \sqrt{3} - \sqrt{2}} - \frac{1}{t + \sqrt{3} + \sqrt{2}}\right] \mathrm{dt}$

$= \frac{1}{6 \sqrt{2}} \left[\ln | t + \sqrt{3} - \sqrt{2} | - \ln | t + \sqrt{3} + \sqrt{2} |\right] + c$

$\therefore I = \frac{1}{6 \sqrt{2}} \left\{\ln | \frac{t + \sqrt{3} - \sqrt{2}}{t + \sqrt{3} + \sqrt{2}} |\right\} + c \ldots \to \left(1\right)$

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 $\sin u = \frac{{e}^{x} - 6}{\sqrt{108}}$

$t = \frac{1 - \sqrt{1 - {\left(\frac{{e}^{x} - 6}{\sqrt{108}}\right)}^{2}}}{\frac{{e}^{x} - 6}{\sqrt{108}}} \implies \textcolor{v i o \le t}{t = \frac{\sqrt{108} - \sqrt{108 - {\left({e}^{x} - 6\right)}^{2}}}{{e}^{x} - 6}}$

$\therefore t = \frac{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72}}{{e}^{x} - 6}$

Hence ,from $\left(1\right)$

$I = \frac{1}{6 \sqrt{2}} \left\{\ln | \frac{\textcolor{v i o \le t}{\frac{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72}}{{e}^{x} - 6}} + \sqrt{3} - \sqrt{2}}{\textcolor{v i o \le t}{\frac{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72}}{{e}^{x} - 6}} + \sqrt{3} + \sqrt{2}} |\right\} + c$

$I$=$\frac{1}{6 \sqrt{2}} \ln | \frac{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72} + \left(\sqrt{3} - \sqrt{2}\right) \left({e}^{x} - 6\right)}{6 \sqrt{3} - \sqrt{- {e}^{2 x} + 12 x + 72} + \left(\sqrt{3} + \sqrt{2}\right) \left({e}^{x} - 6\right)} |$$+ c$

$\frac{1}{6 \setminus \sqrt{2}} \setminus \ln | \setminus \frac{{e}^{x} \left(\setminus \sqrt{3} - \setminus \sqrt{2}\right) - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72} + 6 \setminus \sqrt{2}}{{e}^{x} \left(\setminus \sqrt{3} + \setminus \sqrt{2}\right) - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72} - 6 \setminus \sqrt{2}} | + C$

#### Explanation:

Given that

$\setminus \int \setminus \frac{1}{\setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72}} \setminus \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{108 - \left({\left({e}^{x}\right)}^{2} - 2 \left(6\right) {e}^{x} + {6}^{2}\right)}} \setminus \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{108 - {\left({e}^{x} - 6\right)}^{2}}} \setminus \mathrm{dx}$

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

$\setminus \implies \mathrm{dx} = \setminus \frac{6 \setminus \sqrt{3} \setminus \cos \setminus \theta \setminus d \setminus \theta}{6 \setminus \sqrt{3} \setminus \sin \setminus \theta + 6}$

$\setminus \therefore \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{108 - {\left({e}^{x} - 6\right)}^{2}}}$

$= \setminus \int \setminus \frac{\setminus \frac{6 \setminus \sqrt{3} \setminus \cos \setminus \theta \setminus d \setminus \theta}{6 \setminus \sqrt{3} \setminus \sin \setminus \theta + 6}}{\setminus \sqrt{108 - 108 \setminus {\sin}^{2} \setminus \theta}}$

$= \setminus \int \setminus \frac{6 \setminus \sqrt{3} \setminus \cos \setminus \theta \setminus d \setminus \theta}{6 \setminus \sqrt{3} \cos \setminus \theta \left(6 \setminus \sqrt{3} \setminus \sin \setminus \theta + 6\right)}$

$= \setminus \int \setminus \frac{d \setminus \theta}{6 \setminus \sqrt{3} \setminus \sin \setminus \theta + 6}$

$= \setminus \int \setminus \frac{d \setminus \theta}{6 \setminus \sqrt{3} \left(\setminus \frac{2 \setminus \tan \left(\setminus \frac{\theta}{2}\right)}{1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)}\right) + 6}$

$= \setminus \int \setminus \frac{\left(1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)\right) d \setminus \theta}{12 \setminus \sqrt{3} \setminus \tan \left(\setminus \frac{\theta}{2}\right) + 6 \left(1 + \setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right)\right)}$

$= \frac{1}{6} \setminus \int \setminus \frac{{\sec}^{2} \left(\setminus \frac{\theta}{2}\right) d \setminus \theta}{\setminus {\tan}^{2} \left(\setminus \frac{\theta}{2}\right) + 2 \setminus \sqrt{3} \setminus \setminus \tan \left(\setminus \frac{\theta}{2}\right) + 1}$

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

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

Using standard formula: $\setminus \int \setminus \frac{\mathrm{dt}}{{t}^{2} - {a}^{2}} = \frac{1}{2 a} \setminus \ln | \setminus \frac{t - a}{t + a} |$

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

$= \frac{1}{6 \setminus \sqrt{2}} \setminus \ln | \setminus \frac{\setminus \frac{6 \setminus \sqrt{3} - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72}}{{e}^{x} - 6} + \setminus \sqrt{3} - \setminus \sqrt{2}}{\setminus \frac{6 \setminus \sqrt{3} - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72}}{{e}^{x} - 6} + \setminus \sqrt{3} + \setminus \sqrt{2}} | + C$

$= \frac{1}{6 \setminus \sqrt{2}} \setminus \ln | \setminus \frac{{e}^{x} \left(\setminus \sqrt{3} - \setminus \sqrt{2}\right) - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72} + 6 \setminus \sqrt{2}}{{e}^{x} \left(\setminus \sqrt{3} + \setminus \sqrt{2}\right) - \setminus \sqrt{- {e}^{2 x} + 12 {e}^{x} + 72} - 6 \setminus \sqrt{2}} | + C$