问一道不定积分的题目.纠结了好久根号下1-cosx的不定积分.

Question

问一道不定积分的题目.纠结了好久
根号下1-cosx 的不定积分.

Answer 1

令t = 1 - cosx,dt = sinx dx = √(1 - cos²x) dx = √[1 - (1 - t)²] dx = √t√(2 - t) dx
dx = dt/[√t√(2 - t)]
∫ √(1 - cosx) dx
= ∫ √t • dt/[√t√(2 - t)]
= ∫ dt/√(2 - t)
= 2√(2 - t) + C
= 2√[2 - (1 - cosx)] + C
= 2√(1 + cosx) + C
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∫ √(1 - cosx) dx
= ∫ √[2sin²(x/2)] dx
= √2∫ |sin(x/2)| dx,|sin(x/2)|周期2π
当x∈[4(k - 1)π,2(2k - 1)π]
积分 = √2∫ sin(x/2) dx
= 2√2(-cos(x/2)) + C
= -2√2cos(x/2) + C
当x∈[2(2k - 1)π,4kπ]
积分 = √2∫ -sin(x/2) dx
= -2√2(-cos(x/2)) + C
= 2√2cos(x/2) + C