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Easy

Question

If 2 not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent x d x equals not stretchy integral subscript 0 end subscript superscript 1 end superscript c o t to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x, then not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x is equal to

  1. fraction numerator pi over denominator 2 end fraction plus l o g 2    
  2. log2    
  3. fraction numerator pi over denominator 2 end fraction minus l o g 4    
  4. log4    

The correct answer is: log2


    We have
    2 not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent x d x equals not stretchy integral subscript 0 end subscript superscript 1 end superscript c o t to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x
    equals not stretchy integral subscript 0 end subscript superscript 1 end superscript left parenthesis fraction numerator pi over denominator 2 end fraction minus t a n to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis right parenthesis d x
    equals fraction numerator pi over denominator 2 end fraction minus not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x
    Therefore,
    not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x equals fraction numerator pi over denominator 2 end fraction minus 2 not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent d x equals fraction numerator pi over denominator 2 end fraction minus 2 l
    rightwards double arrow l equals not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent x d x equals left parenthesis x right parenthesis t a n to the power of negative 1 end exponent x vertical line subscript 0 end subscript superscript 1 end superscript minus fraction numerator 1 over denominator 2 end fraction not stretchy integral subscript 0 end subscript superscript 1 end superscript fraction numerator 2 x over denominator 1 plus x to the power of 2 end exponent end fraction d x
    equals x t a n to the power of negative 1 end exponent x vertical line subscript 0 end subscript superscript 1 end superscript minus fraction numerator 1 over denominator 2 end fraction I o g left parenthesis 1 plus x to the power of 2 end exponent right parenthesis vertical line subscript 0 end subscript superscript 1 end superscript
    equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction (Iog2‐0) equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction llog2
    Hence,
    not stretchy integral subscript 0 end subscript superscript 1 end superscript t a n to the power of negative 1 end exponent left parenthesis 1 minus x plus x to the power of 2 end exponent right parenthesis d x equals fraction numerator pi over denominator 2 end fraction minus fraction numerator pi over denominator 2 end fraction plus vertical line o g 2 equals vertical line o g 2
    Hence, the correct answer is option (B).

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