求定积分∫上限为π/2下限为0 sin^3/（1+cosx）dx

Question

如题所述

Answer 1

∫ (sin[x])^3 / (1+cos[x]) dx = ∫ -(sin[x])^2 / (1+cos[x]) d(cos[x])
= ∫ ( (cos[x])^2 - 1 ) / (1 + cos[x]) d(cos[x])
= ∫ (cos[x] - 1) * (cos[x] + 1) / (1 + cos[x]) d(cos[x])
= ∫ (cos[x] - 1) d(cos[x]) 换元 cos[x] = t
= ∫ (t - 1) dt
= 1/2 * t^2 - t + C, C为常数.
= 1/2 * (cos[x])^2 - cos[x] + C
∴∫[0,π/2] (sin[x])^3 / (1+cos[x]) dx = ( 1/2 * (cos[x])^2 - cos[x] ) | [0,π/2]
= 1/2.
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