换元法,设x=arcsiny
y=sinx,y=0~1,x=0~π/2,dy=cosxdx,代入:
=∫(0,π/2) πx²cosxdx
= π∫(0, π/2)x²dsinx
= π[x²sinx (0,π/2) 十2∫(0,π/2)xdcosx]
= π[π²/4十2xcosx (0,π/2) -2∫ (0,π/2) cosxdx]
= π[π²/4十2xcosx (0,π/2) -2∫ (0,π/2) cosxdx] =[π²/4 -2∫ (0,π/2) cosxdx]
追答= π[π²/4-2∫ (0,π/2) dsinx]
= π[π²/4-2 sinx(0,π/2)]
=π³/4-2π
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