求∫[0：π/2] xcos2xdx
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解决时间 2021-11-13 15:19
- 提问者网友：一抹荒凉废墟
- 2021-11-13 02:04
求∫[0：π/2] xcos2xdx
最佳答案
- 五星知识达人网友：独行浪子会拥风
- 2021-11-13 03:14
∫(0->π/2) xcos2xdx
=(1/2)∫(0->π/2) xdsin2x
=(1/2)[ x.sin2x]|(0->π/2) -(1/2)∫(0->π/2) sin2x dx
=0-(1/2)∫(0->π/2) sin2x dx
=(1/4)[ cos2x ]|(0->π/2)
=-1/2追问请一问一开头的那个二分之一从哪里来的追答dsin2x =2cos2x dx
∫(0->π/2) xcos2xdx
=(1/2)∫(0->π/2) x(2cos2xdx)
=(1/2)∫(0->π/2) xdsin2x追问我问是这个（∫）前面的二分之从何处来的追答∫(0->π/2) xcos2xdx
=(1/2)∫(0->π/2) x . (2cos2xdx) ( 1 = (1/2)*2 )
=(1/2)∫(0->π/2) xdsin2x
=(1/2)∫(0->π/2) xdsin2x
=(1/2)[ x.sin2x]|(0->π/2) -(1/2)∫(0->π/2) sin2x dx
=0-(1/2)∫(0->π/2) sin2x dx
=(1/4)[ cos2x ]|(0->π/2)
=-1/2追问请一问一开头的那个二分之一从哪里来的追答dsin2x =2cos2x dx
∫(0->π/2) xcos2xdx
=(1/2)∫(0->π/2) x(2cos2xdx)
=(1/2)∫(0->π/2) xdsin2x追问我问是这个（∫）前面的二分之从何处来的追答∫(0->π/2) xcos2xdx
=(1/2)∫(0->π/2) x . (2cos2xdx) ( 1 = (1/2)*2 )
=(1/2)∫(0->π/2) xdsin2x
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