A) \[\frac{\pi }{2}-2\log \sqrt{2}\]
B) \[\frac{\pi }{2}+2\log \sqrt{2}\]
C) \[\frac{\pi }{4}-\log \sqrt{2}\]
D) \[\frac{\pi }{4}+\log \sqrt{2}\]
Correct Answer: A
Solution :
Put \[x=\tan \theta ,\] \[\therefore \] \[dx={{\sec }^{2}}\theta \,d\theta \] As \[x=1\Rightarrow \theta =\frac{\pi }{4}\] and \[x=0\Rightarrow \theta =0\], then \[I=2\int_{0}^{\pi /4}{\theta {{\sec }^{2}}\theta \,d\theta =2[\theta \tan \theta ]_{0}^{\pi /4}-2\int_{0}^{\pi /4}{\tan \theta \,d\theta }}\] = \[\frac{\pi }{2}+2\,[\log \cos x]_{0}^{\pi /4}=\frac{\pi }{2}-2\log \sqrt{2}\].
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