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How to integrate an equation with the square root of a quadratic equation as the denominator?
i need to integrate 3/(12x-6-4x^2)^0.5 but ive no idea what method to use?
1 Answer
- cidyahLv 75 years ago
∫ 3/(12x-6-4x^2)^(1/2)
12x-6-4x^2 = - (4x^2 -12x+6) = - ( (2x)^2 - 2(2x)(3) + (3)^2 - 3^2 +6) = - ( (2x-3)^2 -3) = (3 - (2x-3)^2)
∫ 3/(12x-6-4x^2)^(1/2) = ∫ 3 / ((3-(2x-3)^2 )^(1/2)
∫ 3 / ((3-(2x-3)^2 )^(1/2)
Let u= 2x-3
du = 2 dx
dx = (1/2) du
∫ 3 / ((3-(2x-3)^2 )^(1/2) = (3/2) ∫ du / (3-u^2)^(1/2)
= (3/2) ∫du / [√3(1-u^2/√3)^(1/2)]
= (√3/2) ∫ du / (1 - (u/√3)^2 )^(1/2)
Let s = u/√3
ds = du/ √3
du = √3 ds
= (√3/2) ∫ du / (1 - (u/√3)^2 )^(1/2) = (√3/2)(√3) ∫ ds / (1 - s^2 )^(1/2)
= (3/2) sin^-1(s)
replace s by u/√3
= (3/2) sin^-1(u/√3)
replace u by 2x-3
= (3/2) sin^-1( (2x-3) /√3) +C