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Asked by nitishkrnehu09 7th February 2018, 9:54 PM
Answered by Expert
Answer:
begin mathsize 16px style straight I equals integral subscript 1 over 2014 end subscript superscript 2014 space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x end fraction dx........... left parenthesis straight I right parenthesis
Put space straight x equals 1 over straight y rightwards double arrow dx equals fraction numerator negative 1 over denominator straight y squared end fraction dy
you space get space
straight I equals integral subscript 1 over 2014 end subscript superscript 2014 space fraction numerator cot to the power of negative 1 end exponent space straight y over denominator straight y end fraction dy space space space space space space space space space space space space using space tan to the power of negative 1 end exponent open parentheses 1 over straight y close parentheses equals cot to the power of negative 1 end exponent straight y space for space straight y greater than 0
straight I equals integral subscript 1 over 2014 end subscript superscript 2014 space fraction numerator cot to the power of negative 1 end exponent space straight x over denominator straight x end fraction dx......... left parenthesis II right parenthesis
Adding space left parenthesis straight I right parenthesis space and space left parenthesis II right parenthesis space and space using space tan to the power of negative 1 end exponent straight x plus cot to the power of negative 1 end exponent straight x equals straight pi over 2
Solve space it space further space you space get space straight I equals straight pi over 2 log 2014
space end style
Answered by Expert 8th February 2018, 10:42 AM
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