Find the value of the following integral $$ I = \int_{-\pi}^\pi\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx$$

My attempt: I tried to evaluate this by the property of even/odd functions for definite integrals, but found that the function is neither even nor odd. Further, I found that the function is symmetric with respect to $ x = {\pi}/2 $. But I am not getting an away to proceed to the next step.

Any suggestion will be very helpful.


closed as off-topic by Jack D'Aurizio, Namaste, Glorfindel, user91500, kingW3 Jul 14 '17 at 14:56

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  • 5
    $\begingroup$ Make the substitution $u = -x$ and look at it for a while. $\endgroup$ – Daniel Fischer Jul 14 '17 at 14:19
  • 1
    $\begingroup$ Try substituting $-x$ for $x$. $\endgroup$ – Lord Shark the Unknown Jul 14 '17 at 14:20
  • $\begingroup$ Thanks to @LordSharktheUnknown & DanielFischer. I got the clue. $\endgroup$ – MathsLearner Jul 14 '17 at 14:32

$$ I = \int_{-\pi}^\pi\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx......(1)$$ Substituting $ u = -x $, we get $$ I = \int_{\pi}^{-\pi}\frac{-e^{-\sin u}}{e^{-\sin u}+e^{\sin u}}\,du$$ $$or, I = \int_{-\pi}^{\pi}\frac{e^{-\sin u}}{e^{\sin u}+e^{-\sin u}}\,du$$ $$or, I = \int_{-\pi}^{\pi}\frac{e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}\,du.....(2)$$

Adding (1) & (2) we get,

$$2I = \int_{-\pi}^{\pi}\frac{e^{\sin u}+e^{-\sin u}}{e^{\sin u}+e^{-\sin u}}\,du= 2{\pi}$$ $$or, I = \pi$$


Formally: $$I=\int_{-\pi}^\pi\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx$$ Flipping the domain of integration takes $x$ to $-x$ and preserves the value, while swapping $\sin x$ and $-\sin x$: $$I=\int_{-\pi}^\pi\frac{e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx$$ Add together: $$2I=\int_{-\pi}^\pi\frac{e^{\sin x}+e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx=\int_{-\pi}^\pi1\,dx=2\pi$$ and thus $I=\frac{2\pi}2=\pi$.


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