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Integrals

Question
CBSEENMA12032557

Evaluate: integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus tan to the power of 5 straight x end fraction dx

Solution

Let I = straight I space equals integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus tan to the power of 5 straight x end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus begin display style fraction numerator sin to the power of 5 straight x over denominator cos to the power of 5 straight x end fraction end style end fraction dx
therefore space space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 straight x over denominator cos to the power of 5 straight x plus sin to the power of 5 straight x end fraction dx                                 ...(1)
therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus sin to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction space open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis 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
                             ...(2)
Adding (1) and (2), we get,
therefore space space space space space space space space space space 2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript open parentheses fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction plus fraction numerator sin to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction close parentheses dx
                   space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin to the power of 5 straight x plus cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript space 1 space dx space space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space straight pi over 2 minus 0
therefore space space space space space 2 space space space straight I space space equals space straight pi over 2 space space space space space rightwards double arrow space space space space straight I space equals space straight pi over 4
                   

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