A) \[\pi -\log 2\]
B) \[\pi +\log 2\]
C) \[\frac{\pi }{2}+\log 2\]
D) \[\frac{\pi }{2}-\log 2\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{1}{{{\cot }^{-1}}\,(1-x+{{x}^{2}})dx}\] \[=\int_{0}^{1}{{{\tan }^{-1}}}\left( \frac{1}{{{x}^{2}}-x+1} \right)dx\] \[=\int_{0}^{1}{{{\tan }^{-1}}}\left( \frac{x-(x-1)}{1+x(x-1)} \right)dx\] \[=\int_{0}^{1}{{{\tan }^{-1}}}\,\,x\,dx-\int_{0}^{1}{{{\tan }^{-1}}}(x-1)dx\] \[=\int_{0}^{1}{{{\tan }^{-1}}}\,\,x\,dx-\int_{0}^{1}{{{\tan }^{-1}}}(1-x-1)dx\] \[=2\int_{0}^{1}{{{\tan }^{-1}}\,x\,dx=2\left[ x\,{{\tan }^{-1}}x-\int{\frac{x}{1+{{x}^{2}}}dx} \right]_{0}^{1}}\] \[=2\left[ x\,\,{{\tan }^{-1}}\,x-\frac{1}{2}\,\log (1+{{x}^{2}}) \right]_{0}^{1}\] \[=2\left[ \left( 1\,{{\tan }^{-1}}1-\frac{1}{2}\log \,2 \right)-\left( 0-\frac{1}{2}\,\log 1 \right) \right]\] \[=\frac{\pi }{2}-\log 2\]
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