Question

Y乘根号下(2y-y平方）怎么积分？

Answer 1

设该积分为J，则
J=∫y√(2y-y²)dy
=∫y√[1-(1-2y+y²)]dy
=∫√y√[1-(1-y)²)]dy
设1-y=cost，则
y=1+cost，dy=-sintdt
原式变为
J=∫[(1+cost)sint(-sint)]dt
=-∫(sin²t)(1+cost)dt
=-∫(sin²t)dt-∫(sin²tcost)dt
=-∫(1/2)(1-cos2t)dt-∫(sin²t)d(sint)
=-t/2+sin2t/4-sin³t/3+C
其中，t=arccos(1-y)，
sint=√(2y-y²)，
sin2t=2sintcost=2(1-y)√(2y-y²)

Answer 2

∫y√(2y-y^2)dy
令y=2(sinx)^2,dy=4sinxcosxdx
原式=∫2(sinx)^2{√2(sinx)^2[2-2(sinx)^2]}4sinxcosxdx
=16∫(sinx)^4(cosx)^2dx
=4∫(sin2x)^2(sinx)^2dx
=2∫(sin2x)^2(1-cos2x)dx
=2∫(sin2x)^2dx-2∫(sin2x)^2cos2xdx
=∫(1-cos4x)dx-∫(sin2x)^2dsin2x
=x - (sin4x)/4 - [(sin2x)^3]/3 +C