Question

∫x√（2x-x∧2）dx

Answer 1

令x-1=sinu，则x=1+sinu，u=arcsin(x-1)
∫x√(2x-x²)dx
=∫x√[1-(x-1)²]d(x-1)
=∫(1+sinu)√(1-sin²u)d(sinu)
=∫(1+sinu)·cos²udu
=½∫(1+cos2u)du-∫cos²ud(cosu)
=½u+½sinucosu-⅓cos³u+C
=½arcsin(x-1)+½(x-1)√(2x-x²)-⅓(2x-x²)√(2x-x²)+C
=(1/6)[(2x²-x-3)√(2x-x²)+3arcsin(x-1)]+C

Answer 2

三角换元脱根号，
换元x=1+sinu，
=∫(1+sinu)cosud(1+sinu)
=∫cos²u+sinucos²udu
=1/2∫cos2u+1du-∫cos²udcosu
=sin2u/4+u/2-cos³u/3+C