int_{frac{-pi}{2}}^{frac{pi}{2}} frac{cos x}{ | vedclass

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(A) Let $I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)

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(A) Let $I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(\frac{-\pi}{2} + \frac{\pi}{2} - x)}{1+e^{(\frac{-\pi}{2} + \frac{\pi}{2} - x)}} dx$$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1+e^{-x}} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\frac{1}{e^x}} dx$$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{e^x+1} dx$ ... $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x + e^x \cos x}{1+e^x} dx$$2I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x(1+e^x)}{1+e^x} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x dx$Since $\cos x$ is an even function,$\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$:$2I = 2\int_{0}^{\frac{\pi}{2}} \cos x dx$$I = \int_{0}^{\frac{\pi}{2}} \cos x dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$.

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x=$

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