The value of the definite integral $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{\left(1+e^{x\cos x}\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}$

$\begin{array}{rcl}\mathrm{I}& =& {\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{\mathrm{dx}}{\left(1+{\mathrm{e}}^{\mathrm{x} \cos \mathrm{x}}\right) \left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}\right)}\\ & =& {\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{\mathrm{dx}}{\left(1+{\mathrm{e}}^{-\mathrm{x} \cos \mathrm{x}}\right) \left({\sin }^4 (-\mathrm{x})+{\cos }^4(-1)\right)}\\ & =& {\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{\mathrm{dx}.{\mathrm{e}}^{\mathrm{x} \cos \mathrm{x}}}{\left({\mathrm{e}}^{\mathrm{x} \cos \mathrm{x}}+1\right) \left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}\right)}\end{array}$

$\begin{array}{rcl}2\mathrm{I}& =& ={\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{1+{\mathrm{e}}^{\mathrm{x} \cos \mathrm{x}}}{\left(1+{\mathrm{e}}^{\mathrm{x} \cos \mathrm{x}}\right) \left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}\right)}\\ 2\mathrm{I}& =& ={\int }_{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{{\sec }^4\mathrm{x}}{{\tan }^4\mathrm{x}+1} \mathrm{dx}\\ & =& 2={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{{\sec }^4\mathrm{x}}{{\tan }^4\mathrm{x}+1}\mathrm{dx}\\ \mathrm{I}& =& ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \frac{\left(1+{\tan }^2\mathrm{x}\right) {\sec }^2\mathrm{x}}{{\tan }^4\mathrm{x}+1} \mathrm{dx} \\ & & \end{array}$

$$\begin{gathered} \mathrm{Put} \tan \mathrm{x}=\mathrm{t} \\ {\sec }^2\mathrm{x} \mathrm{dx}=\mathrm{dt} \\ =={\int }_0^1 \frac{\left(1+{\mathrm{t}}^2\right)}{{\mathrm{t}}^4+1}\mathrm{dt} \\ ={\int }_0^1 \frac{1+{\displaystyle \frac{1}{{\mathrm{t}}^2}}}{{\mathrm{t}}^2+{\displaystyle \frac{1}{{\mathrm{t}}^2}}} \mathrm{dt} \\ \mathrm{Put} \mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{y} \\ \left(1+\frac{1}{{\mathrm{t}}^2}\right) \mathrm{dt}=\mathrm{dy} \\ =\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\mathrm{y}}{\sqrt{2}} \\ =\frac{1}{\sqrt{2}}{\left({\tan }^{-1}\frac{\left(\mathrm{t}-{\displaystyle \frac{1}{\mathrm{t}}}\right)}{\sqrt{2}}\right)}_0^1 \\ =\frac{\pi }{2\sqrt{2}} \end{gathered}$$