Posted by Katie on Sunday, December 30, 2012 at 7:27pm.
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, Sunday, December 30, 2012 at 7:34pm
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 7:36pm
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, Sunday, December 30, 2012 at 7:55pm
ok thank you very much. Happy New Year
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