A) \[\frac{\pi }{2}\log 2\]
B) \[\pi \log 2\]
C) \[-\pi \log 2\]
D) 0
Correct Answer: C
Solution :
[c] \[I=\int_{0}^{\pi }{\log (1+\cos x)dx=\int_{0}^{\pi }{\log (2\,\,{{\cos }^{2}}\frac{x}{2})dx}}\] \[=\int_{0}^{\pi }{(\log 2+2\log \cos \frac{x}{2})dx}\] \[=\int_{0}^{\pi }{\log 2dx+2\int_{0}^{\pi }{\log \cos \frac{x}{2}dx}}\] \[=\pi \log 2+2\int_{0}^{\pi /2}{(2\,\,\log \,\,\cos \,\,t)dt}\] where \[\frac{x}{2}=t\] \[=\pi \log 2+4\left( -\frac{\pi }{2}\log 2 \right)\] \[=\pi \log 2-2\pi \log 2=-\pi \log 2\] \[\left[ \int_{0}^{\pi /2}{\log \sin \theta d\theta =\int_{0}^{\pi /2}{\log \cos \theta d\theta =-\frac{\pi }{2}\log 2}} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
