A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{8}\]
C) \[\frac{{{\pi }^{2}}}{8}\]
D) \[\frac{{{\pi }^{2}}}{32}\]
Correct Answer: D
Solution :
[d] Let \[I=\int\limits_{0}^{1}{\frac{{{\tan }^{-1}}}{1+{{x}^{2}}}dx}\] Put \[{{\tan }^{-1}}x=t\] \[\frac{1}{1+{{x}^{2}}}dx=dt\] \[x=0,\Rightarrow t=0\] \[x=1,\Rightarrow t=\pi /4\] \[\therefore I=\int\limits_{0}^{\pi /4}{tdt}=\left. \frac{{{t}^{2}}}{2} \right|_{0}^{\pi /4}=\frac{{{\pi }^{2}}}{32}\]
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