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求定积分∫(sinx-cosx)/3次根号下(sinx+cosx) [0,π/2]
人气:223 ℃ 时间:2019-10-31 11:37:49
解答
[0,π/2] ∫(sinx-cosx)/(sinx+cosx)^(1/3) dx
=[0,π/2] ∫-d(sinx+cosx)/(sinx+cosx)^(1/3)
=[0,π/2] ∫-d(sinx+cosx)/(sinx+cosx)^(1/3)
=-3/2 (sinx+cosx)^(2/3)|[0,π/2]
=0=[0,π/2] ∫-d(sinx+cosx)/(sinx+cosx)^(1/3)是什么意思?凑微分,使其与分母中具有相同形式的变量d(sinx+cosx) = (cosx-sinx)dx
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