A) \[a\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{a}\]
D) \[2\pi \]
Correct Answer: B
Solution :
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 \[2I=\int_{-\pi }^{\pi }{\frac{(1+{{a}^{x}})\cos x}{(1+{{a}^{x}})}}dx\] \[\Rightarrow \] \[2I=\int_{-\pi }^{\pi }{{{\cos }^{2}}x}dx\] \[\Rightarrow \] \[2I=\int_{-\pi }^{\pi }{\left( \frac{\cos 2x+1}{2} \right)}dx\] \[\Rightarrow \] \[2I=\frac{1}{2}\left[ \frac{\sin 2x}{2}+x \right]_{-\pi }^{\pi }\] \[\Rightarrow \] \[2I=\frac{1}{2}(\pi +\pi )\] \[\Rightarrow \] \[I=\frac{\pi }{2}\]
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