dx/(2+cosx)上π/2下0的定积分

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dx/(2+cosx)上π/2下0的定积分

dx/(2+cosx)上π/2下0的定积分
dx/(2+cosx)上π/2下0的定积分

dx/(2+cosx)上π/2下0的定积分
积分(0--π/2)dx/(2+cosx)
=(0--π/2)(2√3/3)arctan[(√3/3)tan(x/2)
=(2√3/3)arctan(√6/6)

楼上seems not right !,π/(3√3) should be the correct one !
令u = tan(x/2),dx = 2du/(1 + u²),cosx = (1 - u²)/(1 + u²)
当x = 0,u = 0
当x = π/2,u = 1
∫(0~π/2) dx/(2 + cosx)
=...

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楼上seems not right !,π/(3√3) should be the correct one !
令u = tan(x/2),dx = 2du/(1 + u²),cosx = (1 - u²)/(1 + u²)
当x = 0,u = 0
当x = π/2,u = 1
∫(0~π/2) dx/(2 + cosx)
= ∫(0~1) 1/[2 + (1 - u²)/(1 + u²)] * 2/(1 + u²) du
= ∫(0~1) (1 + u²)/(2u² + 2 + 1 - u²) * 2/(1 + u²) du
= ∫(0~1) 2/(3 + u²) du
= 2/√3 * arctan(u/√3) |(0~1)
= 2/√3 * arctan(1/√3)
= 2/√3 * π/6
= π/(3√3)

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