Official solution
\[I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} \] .... (1)
Using \[\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } \]
\[I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{ - x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} \]
Add (1) and (2)
\[2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}} \]
\[2I = 2\int\limits_0^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}} \]
\[I = \int\limits_0^{{\pi \over 4}} {{{\left( {1 + {1 \over {{{\tan }^2}x}}} \right){{\sec }^2}x} \over {{{\left( {\tan - {1 \over {\tan x}}} \right)}^2} + 2}}} dx\]
\[\tan x - {1 \over {\tan x}} = t\]
\[\left( {1 + {1 \over {{{\tan }^2}x}}} \right){\sec ^2}xdx = dt\]
\[I = \int\limits_{ - \infty }^0 {{{dt} \over {{t^2} + 2}}} = \left[ {{1 \over {\sqrt 2 }}{{\tan }^{ - 1}}\left( {{t \over {\sqrt 2 }}} \right)} \right]_{ - \infty }^0\]
\[I = 0 - {1 \over {\sqrt 2 }}\left( { - {\pi \over 2}} \right) = {\pi \over {2\sqrt 2 }}\]