What is meant by local integration?

Change of Variable Integration?

• Integration from -r > r Intrgral of squrt (r^2 - x^2) dx apparently meant to use x=rsint but not sure how r >0

• Answer:

  r ∫ √(r² - x²) dx (r being a positive constant; in cases like this, r usually represents the -r radius of a semicircle whose area is given by the definite integral) let: x = r sinθ sinθ = x/r θ = arcsin(x/r) dx = r cosθ dθ then, substituting: ∫ √(r² - x²) dx = ∫ √[r² - (r sinθ)²] r cosθ dθ = ∫ [√(r² - r²sin²θ)] r cosθ dθ = ∫ {√[r²(1 - sin²θ)]} r cosθ dθ = replace 1 - sin²θ with cos²θ: ∫ [√(r²cos²θ)] r cosθ dθ = ∫ r cosθ r cosθ dθ = (pulling constants out) ∫ r²cos²θ dθ = recall the power-reduction formula cos²θ = [1 + cos(2θ)]/2: ∫ r² {[1 + cos(2θ)]/2} dθ = break it up pulling constants out: (r²/2) ∫ dθ + (r²/2) ∫ cos(2θ) dθ = (r²/2)θ + (r²/2) (1/2)sin(2θ) + C = (according to the double-angle identity sin(2θ) = 2sinθ cosθ) (r²/2)θ + (r²/2) (1/2)2sinθ cosθ + C = (r²/2)θ + (r²/2)sinθ cosθ + C let's now convert this back to x variable, recalling that: θ = arcsin(x/r) sinθ = x/r hence: cosθ = √(1 - sin²θ) = √[1 - (x/r)²] = √[1 - (x²/r²)] = √[(r² - x²) /r²] = [√(r² - x²)] /r and then, substituting back: (r²/2)θ + (r²/2)sinθ cosθ + C = (r²/2)arcsin(x/r) + (r²/2)(x/r) {[√(r² - x²)] /r } + C = (r²/2)arcsin(x/r) + (r²/2)(x/r²)√(r² - x²) + C = ending with: (r²/2)arcsin(x/r) + (1/2)x √(r² - x²) + C (antiderivative) having the antiderivative, plug in the bounds: {(r²/2)arcsin[(r)/r] + (1/2)(r) √[r² - (r)²]} - {(r²/2)arcsin[(- r)/r] + (1/2)(- r) √[r² - (- r)²]} = (r²/2)arcsin(1) + (1/2)r √(r² - r²) - (r²/2)arcsin(- 1) + (1/2)r √(r² - r²) = (r²/2)(π/2) + (1/2)r (0) - (r²/2)(- π/2) + (1/2)r (0) = (r²/2)(π/2) + (r²/2)(π/2) = 2(r²/2)(π/2) = (r²/2)π I hope this helps..