The value of int_{-pi / 2}^{pi / 2} frac{1}{1 | vedclass

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(D) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I

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$\frac{\pi}{2}$

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(D) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin(-\pi / 2 + \pi / 2 - x)}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{-\sin x}} dx$$I = \int_{-\pi / 2}^{\pi / 2} \frac{e^{\sin x}}{e^{\sin x} + 1} dx$  $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{-\pi / 2}^{\pi / 2} \left( \frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx$$2I = \int_{-\pi / 2}^{\pi / 2} \frac{1+e^{\sin x}}{1+e^{\sin x}} dx = \int_{-\pi / 2}^{\pi / 2} 1 dx$$2I = [x]_{-\pi / 2}^{\pi / 2} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$$I = \frac{\pi}{2}$

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+ e^{\sin x}} dx$ is:

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