A) \[-2\pi \,\,ln\,\,2\]
B) \[-\frac{\pi }{4}ln\,\,2\]
C) \[-\pi \,\,ln\,\,2\]
D) \[-\frac{\pi }{2}ln\,\,2\]
Correct Answer: D
Solution :
[d] \[I=\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,\,(\cos \,\,x)}{1+{{e}^{x}}.{{e}^{\sin x}}}dx}\] \[=\int\limits_{-\pi /2}^{\pi /2}{\frac{ln\,\,(\cos \,\,x)}{1+{{e}^{-(x+\sin \,\,x)}}}dx}\] \[\Rightarrow 2I=\int\limits_{-\pi /2}^{\pi /2}{\frac{ln(\cos x)}{1+{{e}^{x+\sin x}}}(1+{{e}^{(x+\sin \,\,x)}})dx}\] \[=\int\limits_{-\pi /2}^{\pi /2}{ln(\cos \,\,x)}dx\]\[2I=2\int\limits_{0}^{\pi /2}{ln(\cos \,\,x)}dx\Rightarrow I=-\frac{\pi }{2}ln\,\,2\]
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