∫π/2到-π/2√cosx-cosx^3dx=

pyxyxf2022-10-04 11:39:541条回答

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50288 共回答了21个问题 | 采纳率95.2%: ∫π/2到-π/2√cosx-cosx^3dx
=-2∫(0,π/2)√cosx（1-cos²x）dx
=-2∫(0,π/2)sinx√cosxdx
=2∫(0,π/2)√cosxdcosx
=2*(2/3)(cosx)^(3/2)|(0,π/2)
=-4/3; 1年前

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