A) \[\frac{\pi }{4}\ln 2\]
B) \[\frac{\pi }{8}\ln 2\]
C) \[\frac{\pi }{16}\ln 2\]
D) \[\frac{\pi }{32}\ln 2\]
Correct Answer: C
Solution :
Let\[I=\int\limits_{0}^{1/2}{\frac{\ln (1+2x)}{1+4{{x}^{2}}}}dx\]or\[\int\limits_{0}^{1/2}{\frac{\ln (1+2x)}{1+{{(2x)}^{2}}}}dx\] Put \[2x=\tan \theta \] \[\therefore \]\[\frac{2dx}{d\theta }={{\sec }^{2}}\theta \]or\[dx=\frac{{{\sec }^{2}}\theta d\theta }{2}\] also when \[x=0\Rightarrow \theta =0\] and when \[x=\frac{1}{2}\Rightarrow \theta ={{45}^{o}}\]or\[\frac{\pi }{4}\] \[\therefore \]\[I=\int\limits_{0}^{\frac{\pi }{4}}{\frac{\ln (1+tan\theta )}{1+{{\tan }^{2}}\theta }}\times \frac{{{\sec }^{2}}\theta d\theta }{2}\] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\frac{\ln (1+tan\theta )}{1+{{\tan }^{2}}\theta }}\times {{\sec }^{2}}\theta d\theta \] \[(\because 1+ta{{n}^{2}}\theta ={{\sec }^{2}}\theta )\] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln (1}+\tan \theta )d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln }\left[ 1+\tan \left( \frac{\pi }{4}-\theta \right) \right]d\theta \] (Using the property of definite integral) \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln }\left[ 1+\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\times \tan \theta } \right]d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln }\left[ 1+\frac{1-\tan \theta }{1+\tan \theta } \right]d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln }\left[ \frac{1+\cancel{\tan \theta }+1-\cancel{\tan \theta }}{1+\tan \theta } \right]d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln }\left[ \frac{2}{1+\tan \theta } \right]d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{[\ln 2-ln(1+tan}\theta )]d\theta \] \[I=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln 2.d}\theta -\frac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\ln (1+tan\theta )d\theta }\] \[\left. I=\frac{1}{2}\ln 2\theta \right|\,\begin{matrix} \pi /4 \\ 0 \\ \end{matrix}-I\](from eq. (1)) \[I+I=\frac{1}{2}\ln 2\left( \frac{\pi }{4}-0 \right)\] \[2I=\frac{1}{2}\times \frac{\pi }{4}\times \ln 2\] \[2I=\frac{1}{8}\ln 2\]or\[I=\frac{\pi }{16}\ln 2\]
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