The value of $$\int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx,\,a > 0,} $$     is-

Solution :
$$\eqalign{ & {\text{Let }}I = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx.....(1)} \cr & \left[ {{\text{Using }}\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } } \right] \cr & = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} .....(2) \cr} $$
Adding equations (1) and (2) we get
$$\eqalign{ & 2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}x\left( {\frac{{1 + {a^x}}}{{1 + {a^x}}}} \right)dx = \int\limits_{ - \pi }^\pi {{{\cos }^2}x\,dx} } \cr & \Rightarrow 2\int\limits_0^\pi {{{\cos }^2}x\,dx} = 2 \times 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\,dx} = 4\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\,dx} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\,dx} = 2\int\limits_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right)\,dx} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {dx - } 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\,dx} \cr & \Rightarrow I + I = 2\left( {\frac{\pi }{2}} \right) = \pi \cr & \Rightarrow I = \frac{\pi }{2} \cr} $$