Calculus II
posted by Katie .
Find x and dx using trigonometric substitution of
(integral) square root of 4x^2  9 divide by x
x = 2/3 sec x
dx = 2/3 tan x dx
Is this right so far?

∫√(4x^29) dx
2x = 3secθ, so
4x^29 = 9sec^2θ9 = 9tan^2θ
x = 2/3 secθ, so
dx = 2/3 secθ tanθ
and you integral becomes
∫3tanθ (2/3 secθ tanθ) dθ
= 2∫secθ tan^2θ dθ
= 2∫(sec^3θ  secθ) dθ
use integration by parts twice for sec^3θ. 
rats. x = 3/2 secθ, so adjust the constant in the following steps.
when you're all done, you can check your answer by typing in the integral at wolframalpha.com 
ok thank you very much. Happy New Year