How do you integrate #int 1/sqrt(e^(2x)-2e^x-24)dx# using trigonometric substitution?
1 Answer
Jun 11, 2018
Use the substitution
Explanation:
Let
#I=int1/sqrt(e^(2x)-2e^x-24)dx#
Rewrite in terms of
#I=inte^-x/sqrt(1-2e^-x-24e^(-2x))dx#
Complete the square in the square root:
#I=sqrt24inte^-x/sqrt(25-(24e^-x+1)^2)dx#
Apply the substitution
#I=sqrt24int1/(5costheta)(-5/24costhetad theta)#
Simplify:
#I=-1/sqrt24intd theta#
The integral is trivial:
#I=-1/sqrt24theta+C#
Reverse the substitution:
#I=-1/sqrt24sin^(-1)((24e^-x+1)/5)+C#