∫（0→a)x^2(a^2-x^2)^(3/2)dx其中a＞0,怎么算,不要只讲思路
2个回答
∫[0,a]x^2√(a^2-x^2)^3dx
=∫[0,a]-√(a^2-x^2)^5dx+∫[0,a]a^2√(x^2-a^2)^3dx
x=asinu x=a,u=π/2 ,x=0,u=0
=∫[0,π/2]-a^6(cosu)^6du+a^4∫[0,π/2]cosu^4du
=(-a^6/8)∫[0,π/2](1+cos2u)^3du+(a^4/4)∫[0,π/2](1+cos2u)^2du
=(-a^6/8)∫[0,π/2] (1+3cos2u+3cos2u^2+cos2u^3)du +(a^4/4)∫[0,π/2](1+2cos2u+cos2u^2)du
=(-a^6/8)(u+(3/2)sin2u+(3/2)(u+sin4u/4)+∫[0,π/2](1/2)(1-sin2u^2)dsin2u) +a^4/4*(u+sin2u+(1/2)(u+sin4u/4)] |[0,π/2]
=(-a^6/8)*(5u/2)+a^4/4*(3u/2) +
(-a^6/8)*(2sin2u)+ (a^4/4)*(sin2u) +
(-a^6/8)*(3/8)(sin4u)+(a4/4)*(1/8)sin4u +
(-a^6/8)*(-1/6)(sin2u)^3 |[0,π/2]
=(-5π/32)a^6+(3π/8)a^4/4
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