if f'(x)=3 - sqrt(4-x^2) what is f(x)?

Hello,

we need to evaluate the indefinite integral:

f(x) = ∫ f '(x) dx =

∫ [3 - √(4 - x²)] dx =

(splitting into two integrals and factoring the constant out)

3 ∫ dx - ∫ √(4 - x²) dx =

3x - ∫ √(4 - x²) dx (##)

let's rewrite the remaining integral as:

∫ √(2² - x²) dx =

let:

x = 2sinθ

dx = 2cosθ dθ

yielding, by trig substitution:

∫ √(2² - x²) dx = ∫ {√[2² - (2sinθ)²]} 2cosθ dθ =

∫ [√(2² - 2²sin²θ)] 2cosθ dθ =

∫ {√[2²(1 - sin²θ)]} 2cosθ dθ =

(applying the identity 1 - sin²θ = cos²θ)

∫ [√(2²cos²θ)] 2cosθ dθ =

∫ (2cosθ) 2cosθ dθ =

∫ 4cos²θ dθ =

(considering the argument θ as (2θ)/2)

∫ 4cos²[(2θ)/2] dθ =

let's apply the half-angle identity cos²(α/2) = (1 + cos α)/2:

∫ 4 {[1 + cos(2θ)] /2} dθ =

∫ 2 [1 + cos(2θ)] dθ =

∫ [2 + 2cos(2θ)] dθ =

(splitting into two integrals)

∫ 2 dθ + ∫ 2cos(2θ) dθ =

2θ + ∫ cos(2θ) 2 dθ =

(being 2, in the remaining integral, the derivative of the argument 2θ)

2θ + ∫ cos(2θ) d(2θ) =

(applying the integration rule ∫ cos[f(x)] d[f(x)] = sin[f(x)] + C)

2θ + sin(2θ) + C =

(applying the double-angle identity sin(2θ) = 2sinθ cosθ)

2θ + 2sinθ cosθ + C

let's now recall that x = 2sinθ; hence:

sinθ = x/2

θ = arcsin(x/2)

cosθ = √(1 - sin²θ) = √[1 - (x/2)²] = √[1 - (x²/4)] = √[(4 - x²)/4] = [√(4 - x²)] /2

then, substituting back:

2θ + 2sinθ cosθ + C = 2arcsin(x/2) + 2(x/2) {[√(4 - x²)] /2} + C =

2arcsin(x/2) + (x/2) √(4 - x²) + C =

2arcsin(x/2) + (1/2)x √(4 - x²) + C

finally let's insert this result into the above (##) expression, yielding:

3x - [2arcsin(x/2) + (1/2)x √(4 - x²)] + C =

3x - 2arcsin(x/2) - (1/2)x √(4 - x²)] + C

in conclusion:

f(x) = 3x - 2arcsin(x/2) - (1/2)x √(4 - x²)] + C

I hope it's helpful