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December 21, 2014

December 21, 2014

Posted by **Katie** on Sunday, December 30, 2012 at 7:27pm.

(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**, Sunday, December 30, 2012 at 7:34pm∫√(4x^2-9) dx

2x = 3secθ, so

4x^2-9 = 9sec^2θ-9 = 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**, Sunday, December 30, 2012 at 7:36pmrats. 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**, Sunday, December 30, 2012 at 7:55pmok thank you very much. Happy New Year

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