∫ (x^2)/sqrt(4 - x^2) dx by letting x = 2sinθ?
∫ x^2/√(4 - x^2) dx => sub. x = 2sin(θ) , dx = 2cos(θ) dθ:
= ∫ [4 sin^2(θ) * 2co(θ)] /√(4 - 4sin^2θ) dθ => simplify
= 4 ∫ sin^2(θ) dθ => sin^2θ = [1 - cos(2θ)]/2
= 2 ∫ [1 - cos(2θ)] dθ
= 2 ∫ dθ - 2 ∫ cos(2θ) dθ
= 2θ - sin(2θ) + C
=> x = 2sin(θ ) => sin(θ) = x/2:
=> cos(θ) = √(1 - sin^2θ) = √(1 - x^2/4) = √(4 - x^2)/2
=> sin(2θ) = 2sin(θ) cos(θ) = 2(x/2)[√(4 - x^2)/2] = x√(4 - x^2)/2
=> sin(θ) = x/2 => θ = arcsin(x/2)
=> sub. back the values:
∫ x^2/√(4 - x^2) dx
= 2arcsin(x/2) - x√(4 - x^2)/2 + C
x = 2sinθ gives dx=x = 2cosθ dθ, 4-x^2=4 - 4sin^2(θ)=4cos^2(θ)
so sqrt(4-x^2)=2cosθ
and integral =∫(4sin^2(θ)2cosθ/(2cosθ) dθ
=∫4sin^2(θ)dθ and from double angel formula
=∫(2-2cos2θ)dθ = 2θ - sin2θ = 2arcsin(x/2) - (1/2)xsqrt(4-x^2)+C
∫ (x^2)/sqrt(4 - x^2) dx by letting x = 2sinθ
∫(4sin^2θ*2cosθ/2cosθ) dθ
∫4sin^2θ dθ
∫2(1-cos2θ) dθ
2θ-sin2θ+c
2sin^-1(x/2) -(x/2)sqrt(4-x^2)+c
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∫ x^2/√(4 - x^2) dx => sub. x = 2sin(θ) , dx = 2cos(θ) dθ:
= ∫ [4 sin^2(θ) * 2co(θ)] /√(4 - 4sin^2θ) dθ => simplify
= 4 ∫ sin^2(θ) dθ => sin^2θ = [1 - cos(2θ)]/2
= 2 ∫ [1 - cos(2θ)] dθ
= 2 ∫ dθ - 2 ∫ cos(2θ) dθ
= 2θ - sin(2θ) + C
=> x = 2sin(θ ) => sin(θ) = x/2:
=> cos(θ) = √(1 - sin^2θ) = √(1 - x^2/4) = √(4 - x^2)/2
=> sin(2θ) = 2sin(θ) cos(θ) = 2(x/2)[√(4 - x^2)/2] = x√(4 - x^2)/2
=> sin(θ) = x/2 => θ = arcsin(x/2)
=> sub. back the values:
∫ x^2/√(4 - x^2) dx
= 2θ - sin(2θ) + C
= 2arcsin(x/2) - x√(4 - x^2)/2 + C
x = 2sinθ gives dx=x = 2cosθ dθ, 4-x^2=4 - 4sin^2(θ)=4cos^2(θ)
so sqrt(4-x^2)=2cosθ
and integral =∫(4sin^2(θ)2cosθ/(2cosθ) dθ
=∫4sin^2(θ)dθ and from double angel formula
=∫(2-2cos2θ)dθ = 2θ - sin2θ = 2arcsin(x/2) - (1/2)xsqrt(4-x^2)+C
∫ (x^2)/sqrt(4 - x^2) dx by letting x = 2sinθ
∫(4sin^2θ*2cosθ/2cosθ) dθ
∫4sin^2θ dθ
∫2(1-cos2θ) dθ
2θ-sin2θ+c
2sin^-1(x/2) -(x/2)sqrt(4-x^2)+c