Question

How do you integrate `int 1/sqrt(e^(2x)+12e^x+40)dx` using trigonometric substitution?

Answer 1

`int \ 1/sqrt(e^(2x)+12e^x+40) \ dx = -sqrt(10)/20 \ "arcsinh" \ (20e^(-x)+3) + C`

Explanation:

We seek:

`I = int \ 1/sqrt(e^(2x)+12e^x+40) \ dx`

`\ \ = int \ 1/sqrt(e^(2x)(1+12e^(-x)+40e^(-2x))) \ dx`

`\ \ = int \ 1/(e^(x)sqrt(1+12e^(-x)+40e^(-2x))) \ dx`

`\ \ = int \ e^(-x)/(sqrt(1+12e^(-x)+40e^(-2x))) \ dx`

We can perform a substitution, Let:

`u = 40e^(-x)+6 => (du)/dx = -40e^(-x)`, and, `e^(-x)=(u-6)/40`

The we can write the integral as:

`I = int \ (-1/40)/(sqrt(1+12((u-6)/40)+40((u-6)/40)^2)) \ du`

`\ \ = -1/40 \ int \ 1/(sqrt(1+12((u-6)/40)+(u-6)^2/40)) \ du`

`\ \ = -1/40 \ int \ 1/(sqrt(1/40(40+12(u-6)+(u-6)^2))) \ du`

`\ \ = -1/40 \ int \ 1/(sqrt(1/40)sqrt(40+12u-72+u^2-12u+36)) \ du`

`\ \ = -sqrt(40)/40 \ int \ 1/(sqrt(4+u^2)) \ du`

`\ \ = -sqrt(40)/40 \ int \ 1/(sqrt(2^2+u^2)) \ du`

This is a standard integral, so we can write:

`I = -sqrt(40)/40 \ "arcsinh" \ (u/2) + C`

And if we restore the substitution, we get:

`I = -sqrt(40)/40 \ "arcsinh" \ ((40e^(-x)+6)/2) + C`

`\ \ = -sqrt(10)/20 \ "arcsinh" \ (20e^(-x)+3) + C`