Question

integrate 4x ^ 2 * sqrt(1 - 4x ^ 2) dx

Answer 1

Let x=½sinθ, dx=½cosθdθ; √(1-4x²)=√(1-sin²θ)=cosθ; 4x²=sin²θ.

Integrand becomes sin²θcosθ(½cosθdθ)=½sin²θcos²θdθ.

Since sin(2θ)≡2sinθcosθ, ⅛sin²(2θ)≡½sin²θcos²θ.

cos(2A)≡1-2sin²A where a is any angle, sin²A≡½(1-cos(2A)).

If A=2θ, sin²(2θ)≡½(1-cos(4θ)).

⅛∫sin²(2θ)dθ=(1/16)∫(1-cos(4θ))dθ=(1/16)[θ-¼sin(4θ)]+C, where C is integration constant.

sin(4θ)=2sin(2θ)cos(2θ)=2(2sinθcosθ)(1-2sin²θ)=

4sinθcosθ(1-2sin²θ)=8x(1-8x²)√(1-4x²).

θ=arcsin(2x), so ∫4x²√(1-4x²)dx=(1/16)[arcsin(2x)-2x(1-8x²)√(1-4x²)]+C.

(arcsin is the same as sin^-1.)