∫π/2到-π/2√cosx-cosx^3dx=
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![∫π/2到-π/2√cosx-cosx^3dx=](/uploads/image/z/12362200-16-0.jpg?t=%E2%88%AB%CF%80%2F2%E5%88%B0-%CF%80%2F2%E2%88%9Acosx-cosx%5E3dx%3D)
∫π/2到-π/2√cosx-cosx^3dx=
∫π/2到-π/2√cosx-cosx^3dx=
∫π/2到-π/2√cosx-cosx^3dx=
∫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