Integral Calculus Question 191
Question: $ \int\limits_{-\pi /2}^{\pi /2}{\frac{ln,(\cos x)}{1+e^{x}.{e^{\sin x}}}dx=} $
Options:
A) $ -2\pi ln2 $
B) $ -\frac{\pi }{4}ln2 $
C) $ -\pi ln2 $
D) $ -\frac{\pi }{2}ln2 $
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Answer:
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}ln2 $
BETA
