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  99
问题
选项
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．
解析
转载请注明原文地址:https://www.kaotiyun.com/show/iIC4777K
0

考研数学三

相关试题推荐
随机试题
最新回复(0)