∫π/2 0 (cos2x/cosx+sinx)dx 的定积分

cooldoir2022-10-04 11:39:542条回答

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寒秋追忆共回答了19个问题 | 采纳率100%: ∫π/2 0 (cos2x/cosx+sinx)dx
=∫π/2 0 (cos²x-sin²x)/(cosx+sinx)dx
=∫π/2 0 (cosx-sinx)dx
=sinx+cosx π/2 0
=(1+0)-(0+1)
=0; 1年前

好运福来共回答了4647个问题 | 采纳率: ∫[0,π/2] (cos2x/cosx+sinx)dx
=∫[0,π/2] (cos^2x-sin^2x)/(cosx+sinx)dx
=∫[0,π/2] (cosx-sinx)dx
=(sinx+cosx)[0,π/2]
=0; 1年前

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f(x)=cos2x/cosx+sinx化简 老跳蚤1年前1: 淘宝2 共回答了14个问题 | 采纳率92.9%

f(x)=cos2x/(sinx+cosx)
=(cos^2x-sin^2x)/(cosx+sinx)
=(cosx+sinx)(cosx-sinx)/(cosx+sinx)
=cosx-sinx
=√2cos(x+π/4)

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