2. 考研
  3. ∫02πf(x-π)dx=∫02πf(x-π)d(x-π)=∫-ππf(z)dx=∫-π0sinx/(1+cos2x)dx+∫0πxsinsbxdx=-arctan(cosx)｜-π0+π/2∫0πsin2xdx=-π/2+π/2×2∫0π/2sin2xd

∫02πf(x-π)dx=∫02πf(x-π)d(x-π)=∫-ππf(z)dx=∫-π0sinx/(1+cos2x)dx+∫0πxsinsbxdx=-arctan(cosx)｜-π0+π/2∫0πsin2xdx=-π/2+π/2×2∫0π/2sin2xd

admin2022-10-25  53
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答案∫02πf(x-π)dx=∫02πf(x-π)d(x-π)=∫-ππf(z)dx=∫-π0sinx/(1+cos2x)dx+∫0πxsinsbxdx=-arctan(cosx)｜-π0+π/2∫0πsin2xdx=-π/2+π/2×2∫0π/2sin2xdx=-π/2+π/2×2×1/2×π/2=-π/2+π2/4．
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考研数学三

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