A) \[\frac{\pi }{2}.log2\]
B) \[\pi .log2\]
C) \[-\pi .log2\]
D) \[-\frac{\pi }{2}.log2\]
Correct Answer: B
Solution :
[b] \[\int\limits_{0}^{\infty }{\log \left( x+\frac{1}{x} \right)}.\frac{dx}{1+{{x}^{2}}}\] Putting \[x=tan\theta \] When \[x=0\therefore \theta \to 0\] When \[x=\infty \therefore \theta \to \frac{\pi }{2}\] Now, \[I=\int\limits_{0}^{\frac{\pi }{2}}{\log (\tan \theta +\cot \theta ).d\theta }=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \right).d\theta }\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{1}{\sin \theta .\cos \theta } \right).d\theta }=\int\limits_{0}^{\frac{\pi }{2}}{\log \left( \frac{2}{2\sin \theta .\cos \theta } \right).d\theta }\] \[=\int\limits_{0}^{\frac{\pi }{2}}{(\log 2-\log \sin 2\theta }).d\theta =[\theta \log 2]_{0}^{\frac{\pi }{2}}-\int\limits_{0}^{\frac{\pi }{2}}{\log \sin 2\theta }.d\theta \] \[=\frac{\pi }{2}.\log 2-{{1}_{1}}\] Now, putting \[2\theta =z\Rightarrow d\theta =\frac{dz}{2}\] \[\therefore {{l}_{1}}\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\log (\sin z).dz}=2.\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\log \sin z.dz}=-\frac{\pi }{2}.\log 2\] \[\therefore {{l}_{1}}\frac{\pi }{2}.\log 2-\left( -\frac{\pi }{2}.\log 2 \right)=\pi .\log 2.\] Hence, option [b] is correct.
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