A) \[\frac{\pi }{4}-\frac{1}{2}{{\log }_{e}}2\]
B) \[\frac{\pi }{2}-{{\log }_{e}}2\]
C) \[\frac{\pi }{2}-\frac{1}{2}{{\log }_{e}}2\]
D) \[\frac{\pi }{4}-{{\log }_{e}}2\]
Correct Answer: A
Solution :
\[I=\int\limits_{0}^{1}{x}\tan \left( \frac{1}{1+{{x}^{2}}({{x}^{2}}-1)} \right)dx\] \[I=\int\limits_{0}^{1}{x}\left( {{\tan }^{-1}}{{x}^{2}}-{{\tan }^{-1}}({{x}^{2}}-1) \right)dx\] \[{{x}^{2}}=t\Rightarrow 2xdx=dt\] \[I=\frac{1}{2}\int\limits_{0}^{1}{\left( {{\tan }^{-1}}t-{{\tan }^{-1}}(t-1) \right)dx}\] \[=\frac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}t\,dt-\frac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}}(t-1)}dt\] \[=\frac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}t\,dt-\frac{1}{2}\int\limits_{0}^{1}{{{\tan }^{-1}}}}dt=\int\limits_{0}^{1}{{{\tan }^{-1}}}dt\] \[{{\tan }^{-1}}t=\theta \Rightarrow t=\tan \theta \] \[dt={{\sec }^{2}}\theta d\theta \] \[\int\limits_{0}^{\pi /4}{\theta .{{\sec }^{2}}}\theta d\theta \] \[I=(\theta .tan\theta )|_{0}^{\pi /4}-\int\limits_{0}^{\pi /4}{\tan \theta }d\theta \] \[\left. =\left( \frac{\pi }{4}-0 \right)-\ln (sec\theta ) \right|_{0}^{\pi /4}\] \[=\left( \frac{\pi }{4} \right)-\left( \ell n\sqrt{2}-0 \right)\] \[=\frac{\pi }{4}-\frac{1}{2}\ell n2\]
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