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

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

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$1$

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(A) Let $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}dx}$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:$I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - \pi /2 + \pi /2 - x)}}{{1 + {e^{ - \pi /2 + \pi /2 - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - x)}}{{1 + {e^{ - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^{ - x}}}}dx$.Since $\cos(-x) = \cos x$,we have $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + 1/e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{e^x + 1}}}dx$.Adding the two expressions for $I$:$2I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^x}}}dx + \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x(1 + e^x)}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\cos x} dx$.$2I = [\sin x]_{ - \pi /2}^{\pi /2} = \sin(\pi /2) - \sin( - \pi /2) = 1 - (-1) = 2$.Therefore,$I = 1$.

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

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