(0至π) ∫ √(1+cos2x) dx
= (0至π) ∫ √(2cos²x) dx
= (0至π) ∫ √2 |cosx| dx
= (0至π/2) ∫ √2 cosx dx + (π/2至π) ∫ -√2 cosx dx
= [ √2 sinx ]| (0至π/2) - [ √2 sinx] | (π/2至π)
= [ √2 sin(π/2) - 0 ] - [ 0 - √2 sin(π/2) ]
= 2√2 sin(π/2)
= 2√2
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