求 ∫ x^3 • cos³x dx

求 ∫ x^3 • cos³x dx
求 ∫ x^3 • cos³x dx

求 ∫ x^3 • cos³x dx
用分部积分法
原式=∫ x^3 • (cosx)^2 • cosxdx
=∫ x^3 • [1-(sinx)^2] d(sinx)
=∫ x^3d(sinx)-∫ x^3 • (sinx)^2d(sinx)
因∫ x^3d(sinx)
=x^3 •sinx-∫ sinxd(x^3)
=x^3 •sinx-3∫ x^2 •sinxdx
=x^3 •sinx+3∫ x^2 d(cosx)
=x^3 •sinx+3[x^2•cosx-∫ cosxd(x^2)]
=x^3 •sinx+3(x^2•cosx-2∫ x •cosxdx)
=x^3 •sinx+3x^2•cosx-6∫ x •d(sinx)
=x^3 •sinx+3x^2•cosx-6(x •sinx-∫ sinxdx)
=x^3 •sinx+3x^2•cosx-6x •sinx+6∫ sinxdx
=x^3 •sinx+3x^2•cosx-6x •sinx-6cosx+C1
又∫ x^3 • (sinx)^2d(sinx)（I）
=(1/3)∫ x^3d[(sinx)^3]
=(1/3)[x^3 •(sinx)^3-∫ (sinx)^3d(x^3)]
=(1/3)x^3 •(sinx)^3-(1/3)∫ (sinx)^3d(x^3)
=(1/3)x^3 •(sinx)^3-∫ x^2•(sinx)^3dx
而∫ x^2•(sinx)^3dx（II）
=∫ x^2•[(cosx)^2-1]d(cosx)
=∫ x^2•(cosx)^2d(cosx)-∫ x^2d(cosx)
其中∫ x^2d(cosx)
=x^2•cosx-∫ cosxd(x^2)
=x^2•cosx-2∫ x•cosxdx
=x^2•cosx-2∫ xd(sinx)
=x^2•cosx-2(x•sinx-∫ sinxdx)
=x^2•cosx-2x•sinx+2∫ sinxdx
=x^2•cosx-2x•sinx-2cosx+C2
其中∫ x^2•(cosx)^2d(cosx)（III）
=(1/3)∫ x^2d[(cosx)^3]
=(1/3)[x^2•(cosx)^3-∫ (cosx)^3d(x^2)]
=(1/3)x^2•(cosx)^3-(2/3)∫ x•(cosx)^3dx
而∫ x•(cosx)^3dx
=∫ x• [1-(sinx)^2] d(sinx)
=∫ xd(sinx)-∫ x• (sinx)^2d(sinx)
其中∫ xd(sinx)
=x• sinx-∫ sinxdx
=x• sinx+cosx+C3
其中∫ x• (sinx)^2d(sinx)
=(1/3)∫ xd [(sinx)^3]
=(1/3)[x•(sinx)^3-∫ (sinx)^3dx]
=(1/3)x•(sinx)^3-(1/3)∫ [(cosx)^2-1]d(cosx)
=(1/3)x•(sinx)^3-(1/3)[∫ (cosx)^2d(cosx)-∫ d(cosx)]
=(1/3)x•(sinx)^3-(1/3)[(1/3)(cosx)^3-cosx]+C4
=(1/3)x•(sinx)^3-(1/9)(cosx)^3+(1/3)cosx+C4
则∫ x•(cosx)^3dx
=(x• sinx+cosx+C3)-[(1/3)x•(sinx)^3-(1/9)(cosx)^3+(1/3)cosx+C4]
=x• sinx+(2/3)cosx-(1/3)x•(sinx)^3+(1/9)(cosx)^3+C5
于是（III）式∫ x^2•(cosx)^2d(cosx)
=(1/3)x^2•(cosx)^3-(2/3)∫ x•(cosx)^3dx
=(1/3)x^2•(cosx)^3-(2/3)[x• sinx+(2/3)cosx-(1/3)x•(sinx)^3+(1/9)(cosx)^3+C5]
=(1/3)x^2•(cosx)^3+(2/3)x•(sinx)^3-(2/3)x• sinx-(2/27)(cosx)^3-(4/9)cosx+C6
于是（II）式∫ x^2•(sinx)^3dx
=[(1/3)x^2•(cosx)^3+(2/3)x•(sinx)^3-(2/3)x• sinx-(2/27)(cosx)^3-(4/9)cosx+C6]-[x^2•cosx-2x•sinx-2cosx+C2]
=(1/3)x^2•(cosx)^3-x^2•cosx+(2/3)x•(sinx)^3+(4/3)x• sinx-(2/27)(cosx)^3+(14/9)cosx+C7
于是（I）式∫ x^3 • (sinx)^2d(sinx)
=(1/3)x^3 •(sinx)^3-(1/3)x^2•(cosx)^3+x^2•cosx-(2/3)x•(sinx)^3-(4/3)x• sinx+(2/27)(cosx)^3-(14/9)cosx+C8
于是原式∫ x^3 • cos³x dx
=x^3 •sinx+3x^2•cosx-6x •sinx-6cosx-(1/3)x^3 •(sinx)^3+(1/3)x^2•(cosx)^3-x^2•cosx+(2/3)x•(sinx)^3+(4/3)x• sinx-(2/27)(cosx)^3+(14/9)cosx+C
=-(1/3)x^3 •(sinx)^3+x^3 •sinx+(1/3)x^2•(cosx)^3+2x^2•cosx+(2/3)x•(sinx)^3-(14/3)x •sinx-(2/27)(cosx)^3-(40/9)cosx+C