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

Question

Charles Anctil · Accepted Answer

#\color{blue}{2/{3\sqrt3} an^{-1}(\frac{2 an (1/2sin^{-1}(\frac{e^x-6}{3}))+1}{\sqrt3})+C# #\int \frac{1}{\sqrt{-e^{2x}+12e^x-27}}\ dx# #=\int \frac{1}{\sqrt{9-(e^{2x}-12e^x+36)}}\ dx# #=\int \frac{1}{\sqrt{9-(e^{x}-6)^2}}\ dx# Let #e^x-6=3\sin heta\implies e^x\ dx=3\cos heta\ d heta# or #dx=\frac{\cos heta\ d heta}{\sin heta+2}#, #=\int \frac{1}{\sqrt{9-9\sin^2 heta}}\ \frac{\cos heta\ d heta}{\sin heta+2}# #=\int \frac{1}{3\cos heta}\ \frac{\cos heta\ d heta}{\sin heta+2}# #=1/3\int \frac{d heta}{\sin heta+2}# #=1/3\int \frac{d heta}{\frac{2 an( heta/2)}{1+ an^2( heta/2)}+2}# #=1/6\int \frac{(1+ an^2( heta/2))\ \ d heta}{ an( heta/2)+1+ an^2( heta/2)}# #=1/3\int \frac{1/2\sec^2( heta/2)\ \ d heta}{( an( heta/2)+1/2)^2+3/4}# #=1/3\int \frac{d( an( heta/2)+1/2)}{( an( heta/2)+1/2)^2+(\sqrt3/2)^2}# Using standard formula: #\int \frac{dt}{t^2+a^2}=1/a an^{-1}(t/a)#, #=1/3 (1/{\sqrt3/2}) an^{-1}(\frac{ an( heta/2)+1/2}{\sqrt3/2})+C# #=2/{3\sqrt3} an^{-1}(\frac{2 an( heta/2)+1}{\sqrt3})+C# #=2/{3\sqrt3} an^{-1}(\frac{2 an (1/2sin^{-1}(\frac{e^x-6}{3}))+1}{\sqrt3})+C#

Paisley Johnson · Answer

#I=2/(3sqrt3)arc tan((e^x-2sqrt(-e^(2x)+12e^x-27))/(sqrt3(e^x-6)))+c#

Here ,

#I=int1/sqrt(-e^(2x)+12e^x-27)dx#

#-e^(2x)+12e^x-27=9-e^(2x)+12e^x-36=9-(e^x-6)^2#

#:.I=int1/sqrt(9-(e^x-6)^2)dx#

Subst. #color(red)(e^x-6=3sinu)=>e^x=6+3sinu=>e^xdx=3cosudu#

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

So,

#I=int1/sqrt(9-9sin^2u)*(3cosudu)/(6+3sinu)#

#=int1/(3cosu) xx (3cosu du)/(6+3sinu)#

#=int1/(6+3sinu)du=1/3int1/(2+sinu)du#

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

#=>du=(2dt)/(1+t^2) and sinu=(2t)/(1+t^2)#

#:.I=1/3int1/(2+(2t)/(1+t^2)) xx(2dt)/(1+t^2)#

#=2/3int1/(2+2t^2+2t)dt=1/3int1/(t^2+t+1)dt#

#=1/3int1/(t^2+t+1/4+3/4)dt=1/3int1/((t+1/2)^2+(sqrt3/2)^2)dt#

#=1/3 1/(sqrt3/2)arc tan((t+1/2)/(sqrt3/2))+c#

#:.I=2/(3sqrt3)arctan((2t+1)/sqrt3)+c......to(A)#

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 #color(red)(sinu=(e^x-6)/3#

#:.t=(1-sqrt(1-((e^x-6)/3)^2))/((e^x-6)/3)#

#:.t=(1-sqrt(9-(e^x-6)^2)/3)/((e^x-6)/3)=(3-sqrt(-e^(2x)+12x-27))/(e^x-6)#

#:.2t+1=(6-2sqrt(-e^(2x)+12e^x-27))/(e^x-6)+1#

#:.2t+1=((6-2sqrt(-e^(2x)+12e^x-27)+e^x-6))/(e^x-6)#

#:.2t+1=(e^x-2sqrt(-e^(2x)+12e^x-27))/(e^x-6)#

So, from #(A)#

#I=2/(3sqrt3)arc tan((e^x-2sqrt(-e^(2x)+12e^x-27))/(sqrt3(e^x-6)))+c#

William Abdalla · Answer

undefined To integrate $ \int \frac{1}{\sqrt{-e^{2x} + 12e^x - 27}} \, dx $ using trigonometric substitution, follow these steps:

1. Recognize the expression $ -e^{2x} + 12e^x - 27 $ as a quadratic in terms of $ e^x $.
2. Rewrite the expression in the form $ a^2 - (e^x)^2 $, where $ a^2 $ is a perfect square.
3. Let $ e^x = a $ and solve for $ dx $.
4. Substitute $ e^x $ with $ a $ and $ dx $ with the corresponding expression.
5. Integrate with respect to $ a $.
6. Finally, substitute back $ e^x $ for $ a $ in your result.