1+cosx=1+(2cos²(x/2)--1]=2cos²(x/2)
0<x/2<π
则0<x<π,cos(x/2)>0
π<x<2π,cos(x/2)<0
所以原式=∫(0,π)|√2cos(x/2)dx+∫(π,2π)|√2cos(x/2)|dx
=√2∫(0,π)cos(x/2)dx-√2∫(π,2π)cos(x/2)dx
=2√2∫(0,π)cos(x/2)d(x/2)-2√2∫(π,2π)cos(x/2)d(x/2)
=2√2[sin(x/2)(0,π)-sin(x/2)(π,2π)]
=2√2[(1-0)-(0-1)]
=4√2
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