Question

问一道不定积分的题目.纠结了好久

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