如何在定义域内求不定积分∫y^2√(4y-y^2)dy

Question

如何在定义域内求不定积分∫y^2√(4y-y^2)dy

Answer 1

∫y^2.√(4y-y^2)dy
consider
4y-y^2 = 4-(2-y)^2
let
2-y = 2sinx
dy =-2cosydx
∫y^2.√(4y-y^2)dy
=∫(2-2sinx)^2( 2cosx)(-2cosy)dx
= -16∫(1-sinx)^2.(sinx.cosx)dx
=-16 ∫ (1- 2sinx+ (sinx)^2 )sinxcosx dx
=-16∫ (sinxcosx)dx - 16 ∫ [-2(sinx)^2 + (sinx)^3] dsinx
=4cos2x -16( -2(sinx)^3/3 + (sinx)^4/4 ] +C
where
2-y = 2sinx