$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $$ ...... (i)

$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $$ ..... (ii)

(i) and (ii)

From equation (i) & (ii)

$$2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}} $$

$$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} } $$

$$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} } $$

when x = 0, t = 0

Now, $$\tan 2x = t$$

when $$x = {\pi \over 4},\,\,t \to \infty $$

$$2{\sec ^2}2x\,dx = dt$$

$$\therefore$$ $$I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty } $$

$$ = 2{\pi \over 2} = \pi $$