∫﹙π/2,0﹚xsin2xdx
=∫﹙π/2,0﹚xsin2xdx
=﹙－1/2﹚∫﹙π/2,0﹚xdcos2x
=﹙－1/2﹚【xcos2x｜﹙π/2,0﹚－∫﹙π/2,0﹚cos2xdx】
=﹙－1/2﹚【﹙－π/2－0﹚－1/2sin2x｜﹙π/2,0﹚】
=π/4+1/4﹙0－0﹚
=π/4
求定积分上限0 下限π/2 xsin2xdx
∫﹙π/2,0﹚xsin2xdx
=∫﹙π/2,0﹚xsin2xdx
=﹙－1/2﹚∫﹙π/2,0﹚xdcos2x
=﹙－1/2﹚【xcos2x｜﹙π/2,0﹚－∫﹙π/2,0﹚cos2xdx】
=﹙－1/2﹚【﹙－π/2－0﹚－1/2sin2x｜﹙π/2,0﹚】
=π/4+1/4﹙0－0﹚
=π/4