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

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(C) Let $I = \int_{ - \pi /2}^{\pi /2} {f(x)dx} $,where $f(x) = \frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}$.Check if $f(x)$ is an odd or

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(C) Let $I = \int_{ - \pi /2}^{\pi /2} {f(x)dx} $,where $f(x) = \frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}$.Check if $f(x)$ is an odd or even function:$f(-x) = \frac{{\sin(-x)}}{{1 + {{\cos }^2}(-x)}}{e^{ - {{\cos }^2}(-x)}}$Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:$f(-x) = \frac{{-\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}} = -f(x)$.Since $f(x)$ is an odd function,by the property of definite integrals $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is odd,we get:$I = 0$.

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

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