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Integrals

Question
CBSEENMA12032665
Wired Faculty App

The value of integral subscript 0 superscript straight pi over 2 end superscript space log space space open parentheses fraction numerator 4 plus 3 space sinx over denominator 4 plus 3 space cosx end fraction close parentheses dx is

  • 2

  • 3 over 4

  • 0

  • -2

Solution

Let
I = integral subscript 0 superscript straight pi over 2 end superscript space log open parentheses fraction numerator 4 plus 3 space sin space straight x over denominator 4 plus 3 space cosx end fraction close parentheses dx                               ...(1)
 equals space integral subscript 0 superscript straight pi log space open curly brackets fraction numerator 4 plus 3 space sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator 4 plus 3 space cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction close curly brackets 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 over 2 end superscript space log space open parentheses fraction numerator 4 plus 3 space cosx over denominator 4 plus 3 space sinx end fraction close parentheses dx
equals space integral subscript 0 superscript straight pi over 2 end superscript space log space open parentheses fraction numerator 4 plus 3 space sinx over denominator 4 plus 3 space cosx end fraction close parentheses to the power of negative 1 end exponent space dx space equals space minus integral subscript 0 superscript straight pi over 2 end superscript log space open parentheses fraction numerator 4 plus 3 space sinx over denominator 4 plus 3 space cosx end fraction close parentheses dx
therefore space space space space space space straight I space equals space minus 1                                                       left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
therefore space space space 2 space straight I space equals space 0 space space space space space space space rightwards double arrow space space space space straight I space equals space 0
therefore space space left parenthesis straight C right parenthesis space is space correct space answer

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