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

Here,

#I=int1/sqrt(-e^(2x)+12e^x+72)dx#

Now,

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

#:.I=int 1/sqrt(108-(e^x-6)^2)dx#

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

#=>e^xdx#=#sqrt108cosudu=>dx=(sqrt108sinu)/(e^x)#=# (sqrt108cosu)/(sqrt108sinu+6)#

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

#I=int1/(sqrt108cosu) xx (sqrt108cosu)/(sqrt108sinu+6)du#

#=int1/(sqrt108sinu+6)du#

#=1/6int1/(sqrt3sinu+1)du#

Substitute , #color(blue)(tan(u/2)=t)=>sec^2(u/2)1/2du=dt#

#=>du=(2dt)/(1+tan^2(u/2))=>color(blue)(du=(2dt)/(1+t^2)) and #

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

So , #I=1/6int1/(sqrt3((2t)/(1+t^2))+1)*(2dt)/(1+t^2)#

#=>I=2/6int1/(2sqrt3t+1+t^2)dt#

#=>I=2/6int1/(t^2+2sqrt3t+3-2)dt to#[completing square]

#=>I=1/3int1/((t+sqrt3)^2-(sqrt2)^2)dt#

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

#=1/(6sqrt2)int[1/(t+sqrt3-sqrt2)-1/(t+sqrt3+sqrt2)]dt#

#=1/(6sqrt2)[ln|t+sqrt3-sqrt2|-ln|t+sqrt3+sqrt2|]+c#

#:.I=1/(6sqrt2){ln|(t+sqrt3-sqrt2)/(t+sqrt3+sqrt2)|}+c...to(1)#

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))#

#=>t^2=(1-cosu)^2/(1-cos^2u)=(1-cosu)^2/sin^2u#

#=>t=(1-cosu)/sinu=(1-sqrt(1-sin^2u))/sinu#

Subst. back #sinu=(e^x-6)/sqrt108#

#t=(1-sqrt(1-((e^x-6)/sqrt108)^2))/((e^x-6)/sqrt108)=>color(violet)(t=(sqrt108-sqrt(108-(e^x-6)^2))/(e^x-6))#

#:. t=(6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6)#

Hence ,from #(1)#

#I=1/(6sqrt2){ln|(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3-sqrt2)/(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3+sqrt2)|}+c#

#I#=#1/(6sqrt2)ln|(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3-sqrt2)(e^x-6))/(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3+sqrt2)(e^x-6))|##+c#

Given that

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

#=\int \frac{1}{\sqrt{108-((e^{x})^2-2(6)e^x+6^2)}}\ dx#

#=\int \frac{1}{\sqrt{108-(e^x-6)^2}}\ dx#

Let #e^x-6=6\sqrt3\sin \theta\implies e^x\ dx=6\sqrt3\cos\theta\ d\theta#

#\implies dx=\frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3\sin\theta+6}#

#\therefore \int \frac{dx}{\sqrt{108-(e^x-6)^2}}#

#=\int \frac{\frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3\sin\theta+6}}{\sqrt{108-108\sin^2\theta}}#

#=\int \frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3cos\theta(6\sqrt3\sin\theta+6)}#

#=\int \frac{ d\theta}{6\sqrt3\sin\theta+6}#

#=\int \frac{ d\theta}{6\sqrt3(\frac{2\tan(\theta/2)}{1+\tan^2(\theta/2)})+6}#

#=\int \frac{ (1+\tan^2(\theta/2))d\theta}{12\sqrt3\tan(\theta/2)+6(1+\tan^2(\theta/2))}#

#=1/{6}\int \frac{ sec^2(\theta/2)d\theta}{\tan^2(\theta/2)+2\sqrt3\ \tan(\theta/2)+1}#

#=1/{3}\int \frac{ 1/2sec^2(\theta/2)d\theta}{(\tan(\theta/2)+\sqrt3)^2-2}#

#=1/{3}\int \frac{ d(\tan(\theta/2)+\sqrt3)}{(\tan(\theta/2)+\sqrt3)^2-(\sqrt2)^2}#

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

#=1/{3}\frac{1}{2\sqrt2}\ln|\frac{\tan(\theta/2)+\sqrt3-\sqrt2}{\tan(\theta/2)+\sqrt3+\sqrt2}|+C#

#=1/{6\sqrt2}\ln|\frac{\frac{6\sqrt3-\sqrt{-e^{2x}+12e^x+72}}{e^x-6}+\sqrt3-\sqrt2}{\frac{6\sqrt3-\sqrt{-e^{2x}+12e^x+72}}{e^x-6}+\sqrt3+\sqrt2}|+C#

#=1/{6\sqrt2}\ln|\frac{e^x(\sqrt3-\sqrt2)-\sqrt{-e^{2x}+12e^x+72}+6\sqrt2}{e^x(\sqrt3+\sqrt2)-\sqrt{-e^{2x}+12e^x+72}-6\sqrt2}|+C#