The value of the definite integral ${\int }_{-\frac{3\mathrm{\pi }}{4}}^{\frac{5\mathrm{\pi }}{4}}\frac{\cos x+\sin x}{1+e^{x-{\displaystyle \frac{\mathrm{\pi }}{4}}}}dx$   equals

$I=\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}\frac{\sin x+\cos x}{1+e^{-\left(x-\frac{\pi }{4}\right)}}dx$  ; $I=\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}\frac{\left(\cos x+\sin x\right)}{1+e^{\left(x-\frac{\pi }{4}\right)}}{e}^{\left(x-\frac{\pi }{4}\right)}dx$
  $\therefore {\left.\underset{\frac{-3\pi }{4}}{\overset{\frac{5\pi }{4}}{\int }}\left(\cos x+\sin x\right)dx=\left(\sin x-\cos x\right)\right]}_{\frac{-3\pi }{4}}^{\frac{5\pi }{4}}=\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-\left(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$
   $2I=0\Rightarrow$ .          $I=0$Ans.