Question

求∫[0：π/2] xcos2xdx

Answer 1

∫(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