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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

  • Calculus II -

    ok thank you very much. Happy New Year

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