Question

求不定积分∫[x^2√(4-x^2)]dx

Answer 1

令x=2sint 则t=arcsinx/2
√4-x^2=2cost ，dx=2costdt
原式=∫4sin^2t4cosx^2tdt
=2∫（1-cos4t）dt
=2t-1/2∫cos4td4t
=2t-sin2tcos2t+c
=2t-2sintcost（1-2sin^2t）+c
=2arcsinxx／2-x√（4-x＾2）／2+x＾3√（4-x＾2）／4+C

Answer 2

∫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)

Answer 3

1/4√(4-x^2) * (x^2-2)+arcssinx/2+c