How do you integrate #int 1/sqrt(e^(2x)-2e^x-24)dx# using trigonometric substitution?

Question

1811 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-11T11:57:07

Use the substitution #24e^-x+1=5sintheta#.

Let

#I=int1/sqrt(e^(2x)-2e^x-24)dx#

Rewrite in terms of #e^-x#:

#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 #24e^-x+1=5sintheta#:

#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#