How do you integrate #int 1/sqrt(e^(2x)-2e^x+2)dx# using trigonometric substitution?
1 Answer
# int \ 1/sqrt(e^(2x)-2e^x+2) \ dx = -sqrt(2)/2 \ "arcsinh" \ (2e^(-x)-1) + C #
Explanation:
We seek:
# I = int \ 1/sqrt(e^(2x)-2e^x+2) \ dx #
# \ \ = int \ 1/sqrt(e^(2x)(1-2e^(-x)+2e^(-2x))) \ dx #
# \ \ = int \ 1/(e^(x)sqrt(1-2e^(-x)+2e^(-2x))) \ dx #
# \ \ = int \ e^(-x)/(sqrt(1-2e^(-x)+2e^(-2x))) \ dx #
We can perform a substitution, Let:
# u = 2e^(-x)-1 => (du)/dx = -2e^(-x) # , and,#e^(-x)=(u+1)/2#
The we can write the integral as:
# I = int \ (-1/2)/(sqrt(1-2((u+1)/2)+2((u+1)/2)^2)) \ du #
# \ \ = -1/2 \ int \ 1/(sqrt(1-2((u+1)/2)+(u+1)^2/2)) \ du #
# \ \ = -1/2 \ int \ 1/(sqrt(1/2(2-2(u+1)+(u+1)^2))) \ du #
# \ \ = -1/2 \ int \ 1/(sqrt(1/2)sqrt(2-2u-2+u^2+2u+1)) \ du #
# \ \ = -sqrt(2)/2 \ int \ 1/(sqrt(u^2+1)) \ du #
This is a standard integral, so we can write:
# I = -sqrt(2)/2 \ "arcsinh" \ (u) + C #
And if we restore the substitution, we get:
# I = -sqrt(2)/2 \ "arcsinh" \ (2e^(-x)-1) + C#
