Trigonometric substitution

[T_TEngineer_AdamT_T]

Hey Guys
I need assistance with this diffucilt problem
\(\displaystyle \L\:\int{\frac{dx}{(4x^2-9)}^{\frac{3}{2}}\)

let x = $\displaystyle 3\sec(\theta)d{\theta}$
let dx = $\displaystyle 3\sec(\theta)\tan(\theta)d{\theta}$

integrating:
\(\displaystyle \int{\frac{dx}{(4(3\sec{\theta})^2-9)^{\frac{3}{2}}}\)
$\displaystyle \int{\frac{dx}{(4(9\sec{\theta})^2-9})^{\frac{3}{2}}}$
\(\displaystyle \int{\frac{dx}{27(4(\sec{\theta})^2-1)^{\frac{3}{2}}}\)

I stop here cause isnt
$\displaystyle 4((\sec{\theta})^{2}-1)$
equal to
$\displaystyle 4(\tan{\theta})^{2}$?

Edit error on latex

 

[mindy88]

4(x^2-9/4)

x=(3/2)sec (x)
dx = (3/2) sec x tan x

the x should be thetas or however you spell it but i can't type that so..

that's how i would start it. i think it's correct.

 

[T_TEngineer_AdamT_T]

hey good idea
thanks alot mindy88!