Question

How do you integrate `int 1/(4x^2-9)^(3/2)` by trigonometric substitution?

Answer 1

`(-x)/(9sqrt(4x^2-9))+C`

Explanation:

`intdx/(4x^2-9)^(3/2)`

Let `2x=3sectheta` such that `2dx=3secthetatanthetad theta`.

`=1/2int(2dx)/((2x)^2-9)^(3/2)`

`=1/2int(3secthetatanthetad theta)/(9sec^2theta-9)^(3/2)`

Factoring `9^(3/2)=27` from the denominator:

`=1/2int(3secthetatantheta)/(27(sec^2theta-1)^(3/2))d theta`

Since `sec^2theta-1=tan^2theta`:

`=1/18int(secthetatantheta)/tan^3thetad theta`

`=1/18intsectheta/tan^2thetad theta`

`=1/18intcostheta/sin^2thetad theta`

Letting `u=sintheta` so `du=costhetad theta`.

`=1/18intu^-2du`

`=-1/(18u)`

`=-1/18csctheta`

`=-1/18sectheta/tantheta`

`=-1/54(3sectheta)/sqrt(sec^2theta-1)`

`=-3/54(3sectheta)/sqrt(9sec^2theta-9)`

Using `3sectheta=2x`:

`=-1/18(2x)/sqrt(4x^2-9)`

`=(-x)/(9sqrt(4x^2-9))+C`