令x=asinu，则sinu=x/a，cosu=√(a²-x²)/a，u=arcsin(x/a)
∫x²√(a²-x²)dx
=∫(asinu)²√(a²-a²sin²u)d(asinu)
=⅛a⁴∫(1-cos4u)du
=a⁴(4u-sin4u)/32 +C
=⅛a⁴(u-sinucosu+2sin³ucosu) +C
=⅛a⁴[arcsin(x/a) -(x/a)·√(a²-x²)/a +2(x/a)³·√(a²-x²)/a] +C
=⅛[a⁴arcsin(x/a) -a²x√(a²-x²) +2x³√(a²-x²)] +C
∫x²√(a²-x²)dx
=∫(asinu)²√(a²-a²sin²u)d(asinu)
=⅛a⁴∫(1-cos4u)du
=a⁴(4u-sin4u)/32 +C
=⅛a⁴(u-sinucosu+2sin³ucosu) +C
=⅛a⁴[arcsin(x/a) -(x/a)·√(a²-x²)/a +2(x/a)³·√(a²-x²)/a] +C
=⅛[a⁴arcsin(x/a) -a²x√(a²-x²) +2x³√(a²-x²)] +C
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