求([(cost)^2]dt 0
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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
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