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CBSE Class 12-science Answered

Evaluate: begin mathsize 18px style integral tan to the power of minus 1 end exponent left parenthesis 1 plus x plus x squared right parenthesis d x end style
Asked by Kaustav Mondal | 27 Sep, 2014, 11:02: PM
answered-by-expert Expert Answer
integral tan to the power of minus 1 end exponent open parentheses 1 plus x plus x squared close parentheses d x equals integral tan to the power of minus 1 end exponent open parentheses fraction numerator 1 plus x left parenthesis 1 plus x right parenthesis over denominator 1 plus x minus x end fraction close parentheses d x equals integral open square brackets straight pi over 2 minus space c o t to the power of minus 1 end exponent space open parentheses fraction numerator 1 plus x left parenthesis 1 plus x right parenthesis over denominator 1 plus x minus x end fraction close parentheses close square brackets d x equals integral straight pi over 2 d x minus space integral c o t to the power of minus 1 end exponent space open parentheses fraction numerator 1 plus x left parenthesis 1 plus x right parenthesis over denominator 1 plus x minus x end fraction close parentheses d x equals πx over 2 minus integral open square brackets c o t to the power of minus 1 end exponent x minus c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis close square brackets d x equals πx over 2 minus integral c o t to the power of minus 1 end exponent x d x plus integral c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis d x equals πx over 2 minus I subscript 1 plus I subscript 2.......... left parenthesis i right parenthesis I subscript 1 equals integral c o t to the power of minus 1 end exponent x. d x equals c o t to the power of minus 1 end exponent x integral d x minus integral open square brackets fraction numerator d left parenthesis c o t to the power of minus 1 end exponent x right parenthesis over denominator d x end fraction integral d x close square brackets d x equals x. c o t to the power of minus 1 end exponent x plus integral fraction numerator x over denominator 1 plus x squared end fraction d x equals x. c o t to the power of minus 1 end exponent x plus 1 half log open vertical bar 1 plus x squared close vertical bar I subscript 2 equals integral c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis d x equals equals c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis integral d x minus integral open square brackets fraction numerator d left parenthesis c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis right parenthesis over denominator d x end fraction integral d x close square brackets d x equals x. c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis plus integral fraction numerator x over denominator 1 plus left parenthesis 1 plus x right parenthesis squared end fraction d x equals x. c o t to the power of minus 1 end exponent x plus 1 half integral fraction numerator 2 x over denominator 1 plus left parenthesis 1 plus x right parenthesis squared end fraction d x equals x. c o t to the power of minus 1 end exponent x plus 1 half integral fraction numerator 2 plus 2 x minus 2 over denominator 1 plus left parenthesis 1 plus x right parenthesis squared end fraction d x equals x. c o t to the power of minus 1 end exponent x plus 1 half integral fraction numerator 2 left parenthesis 1 plus x right parenthesis over denominator 1 plus left parenthesis 1 plus x right parenthesis squared end fraction d x minus 2 over 2 integral fraction numerator 1 over denominator 1 plus left parenthesis 1 plus x right parenthesis squared end fraction d x equals x. c o t to the power of minus 1 end exponent x plus 1 half log open vertical bar 1 plus left parenthesis 1 plus x right parenthesis squared close vertical bar plus c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis S u b s t i t u t i n g space i n space left parenthesis i right parenthesis comma space w e space g e t equals πx over 2 minus x. c o t to the power of minus 1 end exponent x minus 1 half log open vertical bar 1 plus x squared close vertical bar plus x. c o t to the power of minus 1 end exponent x plus 1 half log open vertical bar 1 plus left parenthesis 1 plus x right parenthesis squared close vertical bar plus c o t to the power of minus 1 end exponent left parenthesis 1 plus x right parenthesis plus C
Answered by Prasenjit Paul | 29 Sep, 2014, 11:01: AM

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