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# Evaluate :$\int\limits_0^{2{\pi}}\large\frac{1}{1+e^{sin x}}$$dx This question appeared in the both 65-1 and 65-3 versions of the paper in 2013. Can you answer this question? ## 1 Answer 0 votes Toolbox: • \sin(2\pi-x)=\sin(-x)=-\sin x Given I=\int \limits_0^{2\pi} \large\frac{1}{1+\large e^{\sin x}}$$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 $\sin(2\pi-x)=\sin(-x)=-\sin x$