Question

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 $

Answer 1

Answer: (B)

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} $