Question

求不定积分∫[x^2√(4-x^2)]dx

Answer 1

∫x^2√(4-x^2)dx
=∫(x^2-4)√(4-x^2)dx+4∫√(4-x^2)dx
=∫-√(4-x^2)^3dx+4∫√(4-x^2)dx
= -x√(4-x^2)^3-∫3x^2√(4-x^2)dx+4∫√(4-x^2)dx
4∫x^2√(4-x^2)dx=-x√(4-x^2)^3+4∫√(4-x^2)dx
∫x^2√(4-x^2)dx=(-1/4)x√(4-x^2)^3+∫√(4-x^2)dx
=(-1/4)x√(4-x^2)^3+(1/2)x√(4-x^2)+2arcsin(x/2)+C
∫√(4-x^2)dx=x√(4-x^2)+∫x^2dx/√(4-x^2)=x√(4-x^2)-∫√(4-x^2)dx+4∫dx/√(4-x^2)
2∫√(4-x^2)dx=x√(4-x^2)+4∫d(x/2)/√(1-x^2/4)
∫√(4-x^2)=(1/2)x√(4-x^2)+2arcsin(x/2)