求定积分∫[0,π/2](1+cosx)^1/2dx

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求定积分∫[0,π/2](1+cosx)^1/2dx
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求定积分∫[0,π/2](1+cosx)^1/2dx
求定积分∫[0,π/2](1+cosx)^1/2dx

求定积分∫[0,π/2](1+cosx)^1/2dx

原式=∫[0,π/2](2cos²(x/2))ˆ½dx
=√2∫[0,π/2] |cos(x/2)| dx
=2√2∫[0,π/2] |cos(x/2)| d(x/2)
=2√2sin(x/2)|[0,π∕2]
=2√2(√2/2 - 0)
=2
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