求∫[0:π⼀2] xcos2xdx

2024-11-17 00:55:48
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∫(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