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

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(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$. Using the property $\int_{-a}^{a} f(x) d x = \int_{0}^{a} (f(x) + f(-x)) d x$,we g

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$1$

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(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$. Using the property $\int_{-a}^{a} f(x) d x = \int_{0}^{a} (f(x) + f(-x)) d x$,we get: $I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{\cos(-x)}{1+e^{-x}} \right) d x$. Since $\cos(-x) = \cos x$ and $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,the expression becomes: $I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{e^x \cos x}{1+e^{x}} \right) d x$. $I = \int_{0}^{\pi / 2} \frac{\cos x (1+e^x)}{1+e^x} d x$. $I = \int_{0}^{\pi / 2} \cos x d x$. $I = [\sin x]_{0}^{\pi / 2} = \sin(\pi / 2) - \sin(0) = 1 - 0 = 1$.

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is

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