Question

How do you integrate `int 1/sqrt(-e^(2x)-20e^x-136)dx` using trigonometric substitution?

Answer 1

`int \ 1/sqrt(-e^(2x)-20e^x-136) \ dx = -i/(2sqrt(34)) \ "arcsinh" \ (68/3e^(-x)+5/3) + C`

Explanation:

We seek:

`I = 1/sqrt(-e^(2x)-20e^x-136) \ dx`

`\ \ = int \ 1/sqrt(e^(2x)(-1-20e^(-x)-136e^(-2x))) \ dx`

`\ \ = int \ 1/(e^xsqrt((-1-20e^(-x)-136e^(-2x)))) \ dx`

`\ \ = int \ (e^(-x))/(sqrt(-1-20e^(-x)-136e^(-2x))) \ dx`

We can perform a substitution, Let:

`u = -136e^(-x)-10 => (du)/dx = 136e^(-x)`, and, `e^(-x)=-(u+10)/136`

The we can write the integral as:

`I = int \ ((1/136))/(sqrt(-1+20((u+10)/136)-136((u+10)/136)^2)) \ du`

`\ \ = 1/136 \ int \ (1)/(sqrt(-1 + 20/136(u+10) - 1/136(u+10)^2)) \ du`

`\ \ = 1/136 \ int \ (1)/(sqrt(1/136(-136 + 20(u+10) - (u+10)^2))) \ du`

`\ \ = sqrt(136)/136 \ int \ (1)/(sqrt(-136 +20u+200 - u^2-20u-100)) \ du`

`\ \ = 1/sqrt(136) \ int \ 1/sqrt(-u^2-36) \ du`

`\ \ = 1/(2sqrt(34)) \ int \ 1/sqrt((-1)(u^2+6^2)) \ du`

`\ \ = 1/(2sqrt(34)) \ int \ 1/(isqrt((u^2+6^2))) \ du`

`\ \ = 1/(2sqrt(34)) \ int \ (-i)/(sqrt((u^2+6^2))) \ du`

This is a standard integral, so we can write (temporarily omitting the constant of integration):

`I = -i/(2sqrt(34)) \ "arcsinh" \ (u/6)`

And if we restore the substitution, we get:

`I = -i/(2sqrt(34)) \ "arcsinh" \ ((136e^(-x)+10)/6) + C`

`\ \ = -i/(2sqrt(34)) \ "arcsinh" \ (68/3e^(-x)+5/3) + C`