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Easy

Question

If stretchy integral subscript 0 end subscript superscript 1 end superscript   t a n to the power of negative 1 end exponent invisible function application x d x equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction ln 2 then the value of the definite integral stretchy integral subscript 0 end subscript superscript 1 end superscript   t a n to the power of negative 1 end exponent invisible function application open parentheses 1 minus x plus x to the power of 2 end exponent close parentheses d x equals

  1. Ln 2    
  2. fraction numerator pi over denominator 4 end fraction plus l n invisible function application 2    
  3. fraction numerator pi over denominator 4 end fraction minus l n invisible function application 2    
  4. 2 ln 2    

The correct answer is: Ln 2


    stretchy integral subscript 0 end subscript superscript 1 end superscript   t a n to the power of negative 1 end exponent invisible function application x d x equals fraction numerator pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction l x 2 rightwards double arrow stretchy integral subscript 0 end subscript superscript 1 end superscript   t a n to the power of negative 1 end exponent invisible function application open parentheses 1 minus x plus x to the power of 2 end exponent close parentheses d x
    equals stretchy integral subscript 0 end subscript superscript 1 end superscript   c o t to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 1 minus x plus x to the power of 2 end exponent end fraction close parentheses d x equals stretchy integral subscript 0 end subscript superscript 1 end superscript   fraction numerator pi over denominator 2 end fraction minus open parentheses t a n to the power of negative 1 end exponent invisible function application x plus t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 minus x right parenthesis close parentheses d x
    fraction numerator pi over denominator 2 end fraction minus stretchy integral subscript 0 end subscript superscript 1 end superscript   2 t a n to the power of negative 1 end exponent invisible function application x d x equals fraction numerator pi over denominator 2 end fraction minus open parentheses fraction numerator pi over denominator 2 end fraction minus l x 2 close parentheses equals l x 2

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