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计算定积分∫(上限x,下限-x)(1-X)(sin²x)dx
人气:292 ℃ 时间:2025-03-10 05:39:50
解答

∫[- x→x] (1 - t)sin²t dt
= ∫[- x→x] sin²t dt - ∫[- x→x] tsin²t dt
= 偶函数 + 奇函数
= 2∫[0→x] sin²t dt - 0
= ∫[0→x] (1 - cos2t) dt
= [t - (1/2)sin2t] |[0→x]
= x - (1/2)sin2x
= x - sinxcosx

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