Question

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

Answer 1

`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`