A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[2\pi \]
D) \[\frac{\pi }{6}\]
Correct Answer: A
Solution :
[a] \[I=\int\limits_{0}^{2\pi }{\frac{dx}{1+{{e}^{\sin x}}}}\]????.(i) \[=\int\limits_{0}^{2\pi }{\frac{dx}{1+{{e}^{\sin x(2\pi -x)}}}}=\int\limits_{0}^{2\pi }{\frac{dx}{1+{{e}^{-\sin x}}}}\] \[\Rightarrow I=\int\limits_{0}^{2\pi }{\frac{{{e}^{\sin x}}}{1+{{e}^{\sin x}}}dx}\] ???.(ii) \[\left[ \because \int\limits_{a}^{b}{f(x).dx}=\int\limits_{a}^{b}{f(a+b-x).dx} \right]\] On adding (i) and (ii), we get \[2I=\int\limits_{0}^{2\pi }{dx={{\left[ x \right]}_{0}}^{2\pi }=2\pi }\] Hence, option [a] is correct.
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