Posted by Katie on .
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?

Calculus II 
Steve,
∫√(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. 
Calculus II  oops 
Steve,
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 
Calculus II 
Katie,
ok thank you very much. Happy New Year