# Calculus II

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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 -

∫√(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 -

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 -

ok thank you very much. Happy New Year

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