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CBSE Class 12-science Answered

Evaluate
 
integral subscript negative pi divided by 2 end subscript superscript pi divided by 2 end superscript space cos x space d x divided by space left parenthesis 1 plus e to the power of x right parenthesis
 
 
 
Asked by ahuja8087 | 11 Nov, 2016, 05:56: PM
answered-by-expert Expert Answer
begin mathsize 16px style Let space straight I space equals integral subscript negative straight pi divided by 2 end subscript superscript straight pi divided by 2 end superscript fraction numerator cosx over denominator 1 plus straight e to the power of straight x end fraction dx space space space space space space space space space space equals integral subscript 0 superscript straight pi divided by 2 end superscript open curly brackets fraction numerator cosx over denominator 1 plus straight e to the power of straight x end fraction plus fraction numerator cos left parenthesis negative straight x right parenthesis over denominator 1 plus straight e to the power of negative straight x end exponent end fraction close curly brackets dx space space space..... left square bracket Since space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx equals integral subscript 0 superscript straight a left curly bracket straight f left parenthesis straight x right parenthesis space plus straight f left parenthesis negative straight x right parenthesis right curly bracket dx right square bracket space space space space space space space space space space equals integral subscript 0 superscript straight pi divided by 2 end superscript open curly brackets fraction numerator cosx over denominator 1 plus straight e to the power of straight x end fraction plus fraction numerator cosx over denominator 1 plus straight e to the power of negative straight x end exponent end fraction close curly brackets dx space space space space space space space space space space equals integral subscript 0 superscript straight pi divided by 2 end superscript open curly brackets fraction numerator 1 over denominator 1 plus straight e to the power of straight x end fraction plus fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of straight x end fraction close curly brackets cosx space space dx space space space space space space space space space space equals integral subscript 0 superscript straight pi divided by 2 end superscript open curly brackets fraction numerator 1 plus straight e to the power of straight x over denominator 1 plus straight e to the power of straight x end fraction close curly brackets cosx space space dx space space space space space space space space space space equals integral subscript 0 superscript straight pi divided by 2 end superscript space cosx space space dx space space space space space space space space space space equals space sinx vertical line subscript 0 to the power of straight pi divided by 2 end exponent space space space space space space space space space space equals space sin open parentheses straight pi over 2 close parentheses minus sin 0 equals 1 end style
Answered by Rebecca Fernandes | 09 Feb, 2017, 10:03: AM

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