1/(1-cosx)的不定积分 求~
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1/(1-cosx)的不定积分 求~
![1/(1-cosx)的不定积分 求~](/uploads/image/z/9952560-0-0.jpg?t=1%2F%EF%BC%881-cosx%EF%BC%89%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+%E6%B1%82%EF%BD%9E)
∫ 1/(1 - cosx) dx
= ∫ (1 + cosx)/[(1 - cosx)(1 + cosx)] dx
= ∫ (1 + cosx)/(1 - cos^2(x)) dx
= ∫ (1 + cosx)/sin^2(x) dx
= ∫ (csc^2(x) + cscxcotx) dx
= - cotx - cscx + C
或
∫ 1/(1 - cosx) dx
= ∫ 1/[2sin^2(x/2)] dx
= ∫ csc^2(x/2) d(x/2)
= - cot(x/2) + C
= ∫ (1 + cosx)/[(1 - cosx)(1 + cosx)] dx
= ∫ (1 + cosx)/(1 - cos^2(x)) dx
= ∫ (1 + cosx)/sin^2(x) dx
= ∫ (csc^2(x) + cscxcotx) dx
= - cotx - cscx + C
或
∫ 1/(1 - cosx) dx
= ∫ 1/[2sin^2(x/2)] dx
= ∫ csc^2(x/2) d(x/2)
= - cot(x/2) + C