∫﹙-π,π﹚√﹙1+cos2x)dx =∫﹙-π,π﹚√2|cosx|dx =2∫﹙0,π﹚√2|cosx|dx =2∫﹙0—π/2﹚√2cosxdx -∫(π/2—π﹚√2cosxdx =2√2sinx|(0—π/2)-2√2sinx|(π/2—π)=4√2
求∫(-π,π)根号(1+cos2x)dx
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