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第1个回答 2013-12-21
∫xsin²xdx
=(1/2)∫x(1-cos2x)dx
=(1/2)∫xdx-(1/2)∫ xcos2xdx
=(1/4)x²-(1/4)∫ xd(sin2x)
=(1/4)x²-(1/4)xsin2x+(1/4)∫ (sin2x)dx
=(1/4)x²-(1/4)xsin2x-(1/8)cos2x+C本回答被网友采纳
=(1/2)∫x(1-cos2x)dx
=(1/2)∫xdx-(1/2)∫ xcos2xdx
=(1/4)x²-(1/4)∫ xd(sin2x)
=(1/4)x²-(1/4)xsin2x+(1/4)∫ (sin2x)dx
=(1/4)x²-(1/4)xsin2x-(1/8)cos2x+C本回答被网友采纳
第2个回答 推荐于2021-01-22
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