Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$$ .... (1)

Replace x with $$-$$x,

$$ \therefore $$ $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}} $$ .... (2)

Adding (1) and (2), we get,

$$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx$$

$$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx} $$

$$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx$$

[$${{{\cos }^2}x}$$ is a even function as $$f(x) = f( - x)$$ for $${\cos ^2}x $$]

$$= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx$$

$$ = 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx$$

$$ \Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx$$

$$ = {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}$$

$$ = {1 \over 2}\left[ {{\pi \over 2} - 0} \right]$$

$$ = {\pi \over 4}$$