作业帮0到2a∫x^2*根号(2ax-x^2)dx
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0到2a∫x^2*根号(2ax-x^2)dx
0到2a∫x^2*根号(2ax-x^2)dx
0到2a∫x^2*根号(2ax-x^2)dx
∫x^2√(2ax-x^2)dx
=∫√(2ax-x^2)^3dx+∫2ax√(2ax-x^2)dx
=∫√[(a^2-(x-a)^2]^3dx+2a∫x√(a^2-(a-x)^2)dx
=a^(5/2)∫√(1-(x-a)^2/a^2)^3 d(x/a) +2a^2∫(x-a)√(1-(x-a)^2/a^2) d(x-a) +2a^3∫√(1-(x-a)^2/a^2)d(x-a)
(x-a)/a=sinu 2(x-a)√(2ax-x^2)=a^2sin2u
1-2sinu^2=cos2u=1-2(x-a)^2/a^2=(-a^2-2x^2+4ax)/a^2
sin4u=(x-a)√(2ax-x^2) *(-a^2-2x^2+4ax)/a^4
=a^(5/2)∫cosu^3dsinu+2a^4∫sinucosudsinu+2a^4∫cosudsinu
=a^(5/2)∫cosu^4du+2a^4∫sinucosu^2du+2a^4∫cosu^2du
=a^(5/2)∫[(1+cos2u)^2/4 ]du +2a^4∫sinu-sinu^3du+a^4∫(1+cos2u)du
=a^(5/2) /4 ∫1+2cos2u+(1+cos4u)/2 du -2a^4cosu+2a^4∫(1-cosu^2)dcosu +a^4u+a^4/2 sin2u
=[3a^(5/2)/8+a^4]u +[a^(5/2) /4 +a^4/2 ] sin2u +(a^(5/2)/32) sin4u-(2a^4/3)(cosu)^3+ C
=[3a^(5/2)/8 +a^4]arcsin[(x-a)/a] + [a^(1/2)/2 +a^2] (x-a)√(2ax-x^2) +(a^(5/2)/32) [(x-a)√(2ax-x^2) *(-a^2-2x^2+4ax)/a^4 ] - (2a^4/3) *√[(2ax-x^2) ^3/a^3] +C
∫[0,2a]x^2√(2ax-x^2)=[3a^(5/2) /8 +a^4)*π