\[\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,(\cos x)}{1+{{e}^{x}}.{{e}^{\sin \,\,x}}}dx=}\]

\[\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,(\cos x)}{1+{{e}^{x}}.{{e}^

1.	\[\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,(\cos x)}{1+{{e}^{x}}.{{e}^{\sin \,\,x}}}dx=}\]
A.	\[-2\pi \,\,ln\,\,2\]
B.	\[-\frac{\pi }{4}ln\,\,2\]
C.	\[-\pi \,\,ln\,\,2\]
D.	\[-\frac{\pi }{2}ln\,\,2\]
Answer» E.

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\[\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,(\cos x)}{1+{{e}^{x}}.{{e}^{\sin \,\,x}}}dx=}\]

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