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

Answer

c
(c) $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} $

$\because \frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-{{\cos }^{2}}x}}$  is an odd function, 

$\therefore$ $I = 0$.

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