int_{-pi / 2}^{pi / 2} frac{cos x}{1+e^{x}} d | vedclass

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(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where

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$1$

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(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where $a = -\pi / 2$ and $b = \pi / 2$,we have $a+b = 0$. So,$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos(-x)}{1+e^{-x}} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{-x}} d x$ Multiplying the numerator and denominator by $e^x$,we get: $I = \int_{-\pi / 2}^{\pi / 2} \frac{e^x \cos x}{e^x + 1} d x$ $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x + e^x \cos x}{1+e^x} d x = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x(1+e^x)}{1+e^x} d x$ $2I = \int_{-\pi / 2}^{\pi / 2} \cos x d x$ Since $\cos x$ is an even function,$2I = 2 \int_{0}^{\pi / 2} \cos x d x$ $2I = 2[\sin x]_{0}^{\pi / 2} = 2(1 - 0) = 2$ Therefore,$I = 1$.

$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is equal to

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