The value of $\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ is :

Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$$ .... (1)<br><br>Replace x with $-$x,<br><br>$\therefore$ $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}} $$ .... (2)<br><br>Adding (1) and (2), we get,<br><br>$$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx$$<br><br>$$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx} $$<br><br>$= \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx$<br><br>[${{{\cos }^2}x}$ is a even function as $f(x) = f( - x)$ for ${\cos ^2}x$] <br><br>$= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx$<br><br>$$ = 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx$$<br><br>$\Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx$<br><br>$= {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}$<br><br>$= {1 \over 2}\left[ {{\pi \over 2} - 0} \right]$<br><br>$= {\pi \over 4}$