Asked by Katie
                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?
            
        (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?
Answers
                    Answered by
            Steve
            
    ∫√(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.
    
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.
                    Answered by
            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
    
when you're all done, you can check your answer by typing in the integral at wolframalpha.com
                    Answered by
            Katie
            
    ok thank you very much. Happy New Year
    
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