$\int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$ is equal to

Given integral:

$\int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$

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Let $I = \int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$ ... (1)

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Using the property $\int_{-a}^{a} f(x)dx = \int_{-a}^{a} f(-x)dx$, replace $x$ with $-x$:

$I = \int_{-\pi}^{\pi} \frac{e^{\sin(-x)}}{e^{\sin(-x)} + e^{-\sin(-x)}}dx$

Since $\sin(-x) = -\sin(x)$:

$I = \int_{-\pi}^{\pi} \frac{e^{-\sin x}}{e^{-\sin x} + e^{\sin x}}dx$ ... (2)

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Adding equations (1) and (2):

$I + I = \int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx + \int_{-\pi}^{\pi} \frac{e^{-\sin x}}{e^{-\sin x} + e^{\sin x}}dx$

$2I = \int_{-\pi}^{\pi} \left[\frac{e^{\sin x} + e^{-\sin x}}{e^{\sin x} + e^{-\sin x}}\right]dx$

$2I = \int_{-\pi}^{\pi} 1 \, dx$

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$2I = [x]_{-\pi}^{\pi}$

$2I = \pi - (-\pi)$

$2I = 2\pi$

$I = \pi$

Therefore, the integral equals $\pi$.