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Integrals

Question
CBSEENMA12032501

Evaluate: integral subscript 0 superscript straight pi over 4 end superscript space secx square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction end root dx.

Solution
straight I space equals space integral subscript 0 superscript straight pi over 4 end superscript secx square root of fraction numerator 1 minus sinx over denominator 1 plus sinx end fraction end root dx space equals space integral subscript 0 superscript straight pi over 4 end superscript secx square root of fraction numerator 1 minus sinx over denominator 1 plus sinx end fraction cross times fraction numerator 1 minus sinx over denominator 1 minus sinx end fraction end root dx
        equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator 1 over denominator cos space straight x end fraction. square root of fraction numerator left parenthesis 1 minus sinx right parenthesis squared over denominator 1 minus sin squared straight x end fraction end root dx space equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator 1 over denominator cos space straight x end fraction. square root of fraction numerator 1 minus sinx over denominator cos squared straight x end fraction end root dx
         equals space integral subscript 0 superscript straight pi over 4 end superscript 1 over cosx. fraction numerator 1 minus sinx over denominator cosx end fraction dx space equals space integral subscript 0 superscript straight pi over 4 end superscript fraction numerator 1 minus sinx over denominator cos squared straight x end fraction dx
         
           equals space integral subscript 0 superscript straight pi over 4 end superscript space open parentheses fraction numerator 1 over denominator cos squared straight x end fraction minus 1 over cosx cross times sinx over cosx close parentheses dx space space equals integral subscript 0 superscript straight pi over 4 end superscript left parenthesis sec squared straight x minus secx space tanx right parenthesis space dx
equals space open square brackets tanx space minus space secx close square brackets subscript 0 superscript straight pi over 4 end superscript space equals space open parentheses tan straight pi over 4 minus sec straight pi over 4 close parentheses space minus space left square bracket tan space 0 space minus space sec space 0 right square bracket
equals space open parentheses 1 minus square root of 2 close parentheses space minus space left parenthesis 0 space minus space 1 right parenthesis space equals space 1 minus square root of 2 space plus space 1 space equals space 2 minus square root of 2

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