Question

How do you integrate `int sqrt(e^(8x)-9)` using trig substitutions?

Answer 1

`(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C`

Explanation:

`intsqrt(e^(8x)-9)dx`

Apply the substitution `e^(4x)=3sectheta`. Thus `4e^(4x)dx=3secthetatanthetad theta`.

Note that `dx=(3secthetatanthetad theta)/(4e^(4x))=(3secthetatanthetad theta)/(4(3sectheta))=1/4tanthetad theta`.

`=intsqrt((e^(4x))^2-9)dx`

`=intsqrt(9sec^2theta-9)(1/4tanthetad theta)`

`=3/4intsqrt(sec^2theta-1)(tanthetad theta)`

Note that `tan^2theta=sec^2theta-1`, so:

`=3/4inttan^2thetad theta`

`=3/4int(sec^2theta-1)d theta`

`=3/4intsec^2thetad theta-3/4intd theta`

`=3/4tantheta-3/4theta+C`

From `e^(4x)=3sectheta`, we see that `theta="arcsec"(e^(4x)/3)`. Also, since `sectheta=e^(4x)/3`, we see that the hypotenuse is `e^(4x)`, the adjacent side is `3`, and the opposite side is `sqrt(e^(8x)-9)`. Thus `tantheta=sqrt(e^(8x)-9)/3`.

`=3/4(sqrt(e^(8x)-9)/3)-3/4"arcsec"(e^(4x)/3)+C`

`=(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C`