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cos²x=(1/2) ( 1+cos2x)
偶函数在对称区间上的积分
I = ∫[0,π/2] x² (1+cos2x) dx
∫ x² cos2x dx = (1/2) x² sin2x - ∫ x sin2x dx
= (1/2) x² sin2x + (1/2) x cos2x - (1/2) ∫ cos2x dx
= (1/2) x² sin2x + (1/2) x cos2x - (1/4) sin2x + C
I = [ x³/3 + (1/2) x² sin2x + (1/2) x cos2x - (1/4) sin2x ] | [0,π/2]
= π³/24 - π/4
偶函数在对称区间上的积分
I = ∫[0,π/2] x² (1+cos2x) dx
∫ x² cos2x dx = (1/2) x² sin2x - ∫ x sin2x dx
= (1/2) x² sin2x + (1/2) x cos2x - (1/2) ∫ cos2x dx
= (1/2) x² sin2x + (1/2) x cos2x - (1/4) sin2x + C
I = [ x³/3 + (1/2) x² sin2x + (1/2) x cos2x - (1/4) sin2x ] | [0,π/2]
= π³/24 - π/4
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