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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