• # question_answer The value of$\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}}dx,a>0$is A)  $a\pi$ B)  $\frac{\pi }{2}$ C)  $\frac{\pi }{a}$ D)  $2\pi$

Let$I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}}dx,a>0$           ...(i) $I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}}dx$ (put$x=-x$)     ...(ii) On adding Eqs. (i) and (ii), we get $2I=\int_{-\pi }^{\pi }{\frac{(1+{{a}^{x}})\cos x}{(1+{{a}^{x}})}}dx$ $\Rightarrow$ $2I=\int_{-\pi }^{\pi }{{{\cos }^{2}}x}dx$ $\Rightarrow$ $2I=\int_{-\pi }^{\pi }{\left( \frac{\cos 2x+1}{2} \right)}dx$ $\Rightarrow$ $2I=\frac{1}{2}\left[ \frac{\sin 2x}{2}+x \right]_{-\pi }^{\pi }$ $\Rightarrow$ $2I=\frac{1}{2}(\pi +\pi )$ $\Rightarrow$ $I=\frac{\pi }{2}$