How to integrate an equation with the square root of a quadratic equation as the denominator?

Hello,

∫ [3 /√(12x - 6 - 4x²)] dx =

we need to complete the square inside the root by adding and subtracting 3:

∫ [3 /√(3 + 12x - 3 - 6 - 4x²)] dx =

∫ [3 /√(3 - 4x² + 12x - 9)] dx =

∫ {3 /√[3 - (4x² - 12x + 9)]} dx =

(writing the radicand as a difference of two squares)

∫ {3 /√[(√3)² - (2x - 3)²]} dx =

let:

2x - 3 = (√3)sinθ

(differentiating both sides)

d(2x - 3) = d[(√3)sinθ]

2 dx = (√3)cosθ dθ

dx = (1/2)(√3)cosθ dθ

yielding, by trig substitution:

∫ {3 /√[(√3)² - (2x - 3)²]} dx = ∫ {3 /√{(√3)² - [(√3)sinθ]²} } (1/2)(√3)cosθ dθ =

∫ [3 /√(3 - 3sin²θ)] (1/2)(√3)cosθ dθ =

∫ {3 /√[3(1 - sin²θ)]} (1/2)(√3)cosθ dθ =

(replacing 1 - sin²θ with cos²θ)

∫ [3 /√(3cos²θ)] (1/2)(√3)cosθ dθ =

∫ {3 /[(√3)cosθ]} (1/2)(√3)cosθ dθ =

simplifying to:

∫ 3 (1/2) dθ =

∫ (3/2) dθ =

(3/2)θ + C

let's recall that 2x - 3 = (√3)sinθ

hence:

sinθ = (2x - 3)/√3

θ = arcsin[(2x - 3)/√3]

then, substituting back:

(3/2)θ + C = (3/2)arcsin[(2x - 3)/√3] + C

the answer is:

(3/2)arcsin[(2x - 3)/√3] + C

I hope it's helpful