How do you integrate int 1/sqrt(e^(2x)-2e^x+2)dx∫1√e2x−2ex+2dx 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
