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Integrals

Question
CBSEENMA12032604
Wired Faculty App

By using the properties of definite integrals, evaluate the following integral:
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinxcosx end fraction dx

Solution

Let I = integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinx. cosx end fraction dx                                    ...(1)
therefore space space space space space space space straight I space space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses minus cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator 1 plus sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses space cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction      open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets

rightwards double arrow space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx minus space sinx over denominator 1 plus cosx space sinx end fraction dx space space space rightwards double arrow space space straight I space equals space minus integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinx space cosx end fraction dx rightwards double arrow space space straight I space equals space minus straight I space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets rightwards double arrow space 2 space straight I space equals space 0 space rightwards double arrow space straight I space equals space 0

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