∫(xsinx)²dx怎么算啊,愁- -
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/09/01 08:53:07
∫(xsinx)²dx怎么算啊,愁- -
求达人解,3Q
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
=∫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
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