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Evaluate :$\int\limits_0^{2{\pi}}\large\frac{1}{1+e^{sin x}}$$dx$

This question appeared in the both 65-1 and 65-3 versions of the paper in 2013.
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  • $\sin(2\pi-x)=\sin(-x)=-\sin x$
Given $I=\int \limits_0^{2\pi} \large\frac{1}{1+\large e^{\sin x}}$$dx \quad (1)$
$\textbf{Step 1}$
Applying the properties of integral
$\Rightarrow I=\int \limits_0^{2\pi} \large\frac{1}{1+ \large e^{\sin (2\pi-x)}}dx$
We know that $\sin(2\pi-x)=\sin(-x)=-\sin x$
$\Rightarrow I=\int \limits_0^{2\pi} \large\frac{dx}{1+e^{\sin x}}=\int \limits_0^{2\pi} \large\frac{e^{\sin x}}{1+e^{\sin x}}$$dx \quad (2)$
$\textbf{Step 2}$
Adding $(1)$ and $(2)$, i.e, $\int \limits_0^{2\pi} \large\frac{1}{1+\large e^{\sin x}}$$dx +\int \limits_0^{2\pi} \large\frac{e^{\sin x}}{1+e^{\sin x}}$$dx $
$\Rightarrow 2I=\int \limits_0^{2\pi} \large\frac{1+e^{\sin x}}{1+e^{\sin x}}dx$
$\Rightarrow 2I=\int \limits_0^{2\pi} dx$
$\Rightarrow 2I=\big[x\big]_0^{2\pi}$
$\Rightarrow 2I=2\pi-0 = 2\pi \rightarrow I=\pi$
answered Mar 22, 2013 by balaji.thirumalai

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