这个不定积分怎么求 请问这个不定积分怎么求?
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郭敦顒回答:
(16/3)∫0→π/2 cos^4tdt,
不定积分∫cos^4tdt=(1/4)sin^tcost+(3/4)[t/2-(1/4)sin(2t)]+C
=(1/4)sin^tcost+(3/8)t-(3/16)sin(2t)+C
定积分(16/3)∫0→π/2 cos^4tdt
=[(1/4)sin^3tcost+(3/8)t-(3/16)sin(2t)]| 0→π/2
=(3/16)²π。
降幂再降幂,(cosx)^4 = (1/4)(1+cos2x)^2
= (1/4)[1+2cos2x+(cos2x)^2]
= (1/4)[1+2cos2x+1/2+(1/2)cos4x]
= (1/4)[3/2 + 2cos2x + (1/2)cos4x]
积分得 (1/4)[3x/2 + sin2x + (1/8)sin4x]<0, π/2>
= (1/4)(3π/4) = 3π/16
(16/3)∫ = π
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