求定积分∫(0,π/2) sin²(x/2)dx

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求定积分∫(0,π/2)sin²(x/2)dx求定积分∫(0,π/2)sin²(x/2)dx求定积分∫(0,π/2)sin²(x/2)dx∫(0,π/2)sin²

求定积分∫(0,π/2) sin²(x/2)dx
求定积分∫(0,π/2) sin²(x/2)dx

求定积分∫(0,π/2) sin²(x/2)dx
∫(0,π/2) sin²(x/2)dx
=∫(0,π/2) (1-cosx)/2 dx
=1/2(x-sinx)|(0,π/2)
=π/4-1/2

∫(0,π/2) sin²(x/2)dx
=2∫(0,π/2) sin²(x/2)d(x/2)
=2[-cos(x/2)]|(0,π/2)
=2

求定积分∫(0,π/2) sin²(x/2)dx
解原式=[0,π/2](1/2)∫(1-cosx)dx=[(1/2)(x-sinx)]︱[0,π/2]=π/4-1/2