原式=(1/4)∫(1-cos2x)^2dx =(1/4)∫[1-2cos2x+(cos2x)^2]dx =x/4-(1/4)sin2x+(1/8)∫(1+cos4x)dx =x/4-(1/4)sin2x+x/8+(1/32)sin4x+C =3x/8-(1/4)sin2x+(1/32)sin4x+c.
