∫[0,π/2]sin2x dx
=(1/2)∫[0,π/2]sin2x d(2x)
=-(cos2x)/2 |[0,π/2]
=-(cosπ)/2+(cos0)/2
=1/2+1/2
=1∫[0,π/2]sin2x dx=(1/2)∫[0,π/2]sin2x d(2x)这一步是什么原理这是凑微分 d(2x)=2dx ∴dx=(1/2)d(2x)