求（[(cost)^2]dt 0

求（[(cost)^2]dt 0
求（[(cost)^2]dt 0

求（[(cost)^2]dt 0
[(cost)^2]dt
={[1+cos(2t)]/2}dt
=(1/2)dt+(1/4)cos(2t)d(2t)
=(t/2)+[sin(2t)/4]
因为：0

极限是1