A) \[\pi l\mathbf{og}2\]
B) \[-\pi l\mathbf{og}2\]
C) \[\frac{\pi }{2}l\mathbf{og}2\]
D) \[{{\pi }^{2}}l\mathbf{og}2\]
Correct Answer: B
Solution :
[b] \[I=\int\limits_{0}^{\pi }{log\left( 1+cosx \right).dx}\] ...... (i) \[\Rightarrow I=\int\limits_{0}^{\pi }{log\left( 1+cosx(\pi -x) \right).dx}\] \[\Rightarrow I=\int\limits_{0}^{\pi }{log\left( 1-cosx \right).dx}\] ......(ii) On adding (i) and (ii), we get \[21=\int\limits_{0}^{\pi }{log\left( 1+cosx \right).dx}+\int\limits_{0}^{\pi }{(log\left( 1-cosx \right).dx.}\] \[=\int\limits_{0}^{\pi }{log\left( 1+cosx \right).\left( 1-cosx \right)dx}\] \[=\int\limits_{0}^{\pi }{\log {{\sin }^{2}}x=2.}\int\limits_{0}^{\pi }{\log \sin x=2.}2\int\limits_{0}^{\frac{\pi }{2}}{\log \sin x}\] \[=4\left( -\frac{\pi }{2}.\log 2 \right)\] \[\left[ \therefore \int\limits_{0}^{\frac{\pi }{2}}{\log \sin x.dx=\frac{\pi }{2}\log 2} \right]\] \[\Rightarrow I=-\pi .log2.\] Hence, option [b] is correct.
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