\[\int_{-\,\pi /2}^{\,\pi /2}{\,\frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-{{\cos }^{2}}x}}dx}\] is equal to [AMU 1999]

\[\int_{-\,\pi /2}^{\,\pi /2}{\,\frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-

1.	\[\int_{-\,\pi /2}^{\,\pi /2}{\,\frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-{{\cos }^{2}}x}}dx}\] is equal to [AMU 1999]
A.	\[2{{e}^{-1}}\]
B.	1
C.	0
D.	None of these
Answer» D. None of these

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\[\int_{-\,\pi /2}^{\,\pi /2}{\,\frac{\sin x}{1+{{\cos }^{2}}x}{{e}^{-{{\cos }^{2}}x}}dx}\] is equal to [AMU 1999]

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