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Integrals

Question
CBSEENMA12032607

Show that:
integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space plus space cosx end fraction dx space equals space straight pi squared over 4

Solution

Let I = integral subscript 0 superscript straight pi fraction numerator straight x space tan space straight x over denominator sec space straight x plus cos space straight x end fraction dx
therefore space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tan left parenthesis straight pi minus straight x right parenthesis over denominator sec left parenthesis straight pi minus straight x right parenthesis plus cos left parenthesis straight pi minus straight x right parenthesis end fraction dx     open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets
            equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space left parenthesis negative tanx right parenthesis over denominator negative secx minus cosx end fraction dx space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tanx over denominator secx plus cosx end fraction dx
therefore space space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator straight pi space tanx over denominator sec space straight x space plus space cos space straight x end fraction dx minus integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space plus space tanx end fraction dx
Adding (1) and (2), we get
          2 space straight I space equals integral subscript 0 superscript straight pi fraction numerator straight pi space tanx over denominator secx plus cosx end fraction dx space equals space straight pi integral subscript 0 superscript straight pi fraction numerator tanx over denominator secx plus cosx end fraction dx space equals straight pi integral subscript 0 superscript straight pi fraction numerator begin display style sinx over cosx end style over denominator begin display style 1 over cosx end style plus cosx end fraction dx
therefore space space space space space space space space 2 straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinxdx over denominator 1 plus cos squared straight x end fraction
Put cos x = t, ∴ – sin x dx = dt, or sin x dx = – dt When x = 0, t = cos 0 = 1 When x = straight pi, t = cos straight pi = – 1
therefore space space 2 space straight I space equals space minus straight pi integral subscript 1 superscript negative 1 end superscript fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space straight pi integral subscript negative 1 end subscript superscript 1 fraction numerator dt over denominator 1 plus straight t squared end fraction space equals straight pi open square brackets tan to the power of negative 1 end exponent close square brackets subscript negative 1 end subscript superscript 1 space equals straight pi open square brackets tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space minus tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis close square brackets
space space space space space space space space space space space equals space straight pi open square brackets straight pi over 4 minus open parentheses negative straight pi over 4 close parentheses close square brackets space equals straight pi cross times straight pi over 2
therefore space space space 2 space straight I space equals space straight pi squared over 2 space space                               rightwards double arrow space space space straight I space equals space straight pi over 4

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