∫（xsinx）²dx怎么算啊,愁- -求达人解,3Q

∫（xsinx）²dx怎么算啊,愁- -求达人解,3Q
∫（xsinx）²dx怎么算啊,愁- -
求达人解,3Q

∫（xsinx）²dx怎么算啊,愁- -求达人解,3Q
原式=∫x²sin²xdx
=∫x²(1-cos2x)/2dx
=∫x²/2 dx-∫x²cos2x/2 dx
=x³/6-1/4∫x²cos2xd2x
=x³/6-1/4∫x²dsin2x
=x³/6-1/4[x²sin2x-∫sin2xdx²]
=x³/6-1/4x²sin2x+1/4∫2xsin2xdx
=x³/6-1/4x²sin2x+1/4∫xsin2xd2x
=x³/6-1/4x²sin2x-1/4∫xdcos2x
=x³/6-1/4x²sin2x-1/4xcos2x+1/4∫cos2xd
=x³/6-1/4x²sin2x-1/4xcos2x+1/8sin2x+C

(xsinx)²
=x²(1-cos²x)
=x²-x²(cos2x+1)/2
=x²/2-(x²cos2x)/2
∴∫(xsinx)²dx
=∫x²/2dx-∫(x²cos2x)/2dx
=x³/6-(1/2)(1/2x²sin2...

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(xsinx)²
=x²(1-cos²x)
=x²-x²(cos2x+1)/2
=x²/2-(x²cos2x)/2
∴∫(xsinx)²dx
=∫x²/2dx-∫(x²cos2x)/2dx
=x³/6-(1/2)(1/2x²sin2x-∫xsin2xdx)
=x³/6-(1/4)x²sin2x+(1/2)∫xsin2xdx
至于∫xsin2xdx
=(-1/2)xcos2x-(-1/2))∫cos2xdx
=(-1/2)xcos2x+(1/4)sin2x+c
带回去整理就可以了 楼上真快啊

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∫（xsinx）²dx=(1/2)∫x²(1-cos2x)dx
=(1/2)∫(x²-x²cos2x)dx
=(1/2)[(1/3)x³-∫x²cos2x)dx]
=1/6x³-1/4(x²sin2x-∫2xsin2xdx)
=1/6x³-1/4x²sin2x+1/4...

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∫（xsinx）²dx=(1/2)∫x²(1-cos2x)dx
=(1/2)∫(x²-x²cos2x)dx
=(1/2)[(1/3)x³-∫x²cos2x)dx]
=1/6x³-1/4(x²sin2x-∫2xsin2xdx)
=1/6x³-1/4x²sin2x+1/4(-xcos2x+∫cos2xdx)
=1/6x³-1/4x²sin2x+1/4(-xcos2x+1/2sin2x)+C
=(1/6)x³-(1/4)x²sin2x-(1/4)xcos2x+(1/8)sin2x+C

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