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Integrals

Question
CBSEENMA12032636

Show that:
integral subscript 0 superscript straight pi fraction numerator straight x space sinx over denominator 1 plus cos squared straight x 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 sinx over denominator 1 plus cos squared straight x end fraction dx
therefore 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 sin left parenthesis straight pi minus straight x right parenthesis over denominator 1 plus cos squared left parenthesis straight pi minus straight x right parenthesis end fraction dx space space space space space space space space space space space space space 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 sinx over denominator 1 plus cos squared straight x end fraction dx space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinx over denominator 1 plus cos squared straight x end fraction dx minus integral subscript 0 superscript straight pi fraction numerator straight x space sinx over denominator 1 plus cos squared straight x end fraction dx
 therefore space space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinx over denominator 1 plus cos squared straight x end fraction dx space minus space 1
therefore space space 2 space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sin space straight x space over denominator 1 plus cos squared straight x end fraction dx
Put cos x = t.     ∴ sin x dx = – dt When x = 0, t = cos 0 = 1 When x = straight pi, t = cos straight pi = – 1
therefore space space space 2 space straight I space equals space 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 open square brackets tan to the power of negative 1 end exponent straight t close square brackets subscript 1 superscript negative 1 end superscript
space space space space space space space space space space space space space equals negative straight pi open square brackets tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis space minus space tan to the power of negative 1 end exponent 1 close square brackets space equals space minus straight pi open square brackets negative straight pi over 4 minus straight pi over 4 close square brackets
therefore space space 2 space straight I space equals space straight pi squared over 2 space space rightwards double arrow space space space space straight I space equals space straight pi squared over 4
 

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