Question
Solve this $$\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^x}.{e^{\sin \,x}}}}} dx = ?$$
A.
$$ - 2\pi \,\ln \,2$$
B.
$$ - \frac{\pi }{4}\,\ln \,2$$
C.
$$ - \pi \,\ln \,2$$
D.
$$ - \frac{\pi }{2}\,\ln \,2$$
Answer :
$$ - \frac{\pi }{2}\,\ln \,2$$
Solution :
$$\eqalign{
& I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^x}.{e^{\sin \,x}}}}} dx \cr
& \Rightarrow I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{ - \left( {x + \sin \,x} \right)}}}}} dx \cr
& \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{x + \sin \,x}}}}\left( {1 + {e^{\left( {x + \sin \,x} \right)}}} \right)dx} \cr
& \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr
& \Rightarrow 2I = 2\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr
& \Rightarrow I = - \frac{\pi }{2}\ln \,2 \cr} $$