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Integrals

Question
CBSEENMA12032517

Prove that:
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx space dx over denominator left parenthesis 1 plus sin space straight x right parenthesis space left parenthesis 2 plus sinx right parenthesis end fraction space equals space log space 4 over 3


Solution

Let
straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx space dx over denominator left parenthesis 1 plus sinx right parenthesis space left parenthesis 2 plus sinx right parenthesis end fraction
Put sin x = t,                 ∴ cos x dx = dt 
When x = 0, t = sin 0 = 0
When    straight x space equals space straight pi over 2 comma space space space straight t space equals space sin straight pi over 2 space equals space 1
              I = integral subscript 0 superscript 1 fraction numerator dt over denominator left parenthesis 1 plus straight t right parenthesis space left parenthesis 2 plus straight t right parenthesis end fraction    [Do not forget to change limits of integration]
                 equals space integral subscript 0 superscript 1 open square brackets fraction numerator 1 over denominator left parenthesis 1 plus straight t right parenthesis space left parenthesis 2 minus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 1 minus 2 right parenthesis thin space left parenthesis 2 plus straight t right parenthesis end fraction close square brackets dt
                 equals space integral subscript 0 superscript 1 open square brackets fraction numerator 1 over denominator 1 plus straight t end fraction minus fraction numerator 1 over denominator 2 plus straight t end fraction close square brackets dt space equals space open square brackets log space left parenthesis 1 plus straight t right parenthesis space minus space log left parenthesis 2 plus straight t right parenthesis close square brackets subscript 0 superscript 1
                  equals space open square brackets log space fraction numerator 1 plus straight t over denominator 2 plus straight t end fraction close square brackets subscript 0 superscript 1 space equals space log 2 over 3 minus log 1 half space equals space log open parentheses fraction numerator begin display style 2 over 3 end style over denominator begin display style 1 half end style end fraction close parentheses space equals space log 4 over 3

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