CBSE Class 12-science Answered

Sir Question no-11

Asked by shobhit | 22 Feb, 2019, 11:29: PM

Expert Answer

begin mathsize 16px style straight I equals integral subscript 0 superscript straight pi over 2 end superscript space square root of fraction numerator secx minus tanx over denominator secx plus tanx end fraction end root cross times fraction numerator cosecx over denominator square root of 1 plus 2 cosecx end root end fraction dx straight I equals integral subscript 0 superscript straight pi over 2 end superscript space square root of fraction numerator 1 minus sinx over denominator 1 plus sinx end fraction end root cross times fraction numerator dx over denominator square root of sinx open parentheses 2 plus sinx close parentheses end root end fraction straight I equals integral subscript 0 superscript straight pi over 2 end superscript space square root of fraction numerator open parentheses 1 minus sinx close parentheses open parentheses 1 plus sinx close parentheses over denominator open parentheses 1 plus sinx close parentheses squared end fraction end root cross times fraction numerator dx over denominator square root of sinx open parentheses 2 plus sinx close parentheses end root end fraction straight I equals integral subscript 0 superscript straight pi over 2 end superscript space fraction numerator cosx over denominator 1 plus sinx end fraction cross times fraction numerator dx over denominator square root of sinx left parenthesis 2 plus sinx right parenthesis end root end fraction sinx equals straight t cosxdx equals dt straight x equals 0 space then space straight t equals 0 space and space straight x equals straight pi over 2 space then space straight t equals 1 straight I equals integral subscript 0 superscript 1 space fraction numerator dt over denominator left parenthesis 1 plus straight t right parenthesis square root of straight t open parentheses 2 plus straight t close parentheses end root end fraction Put space log left parenthesis 1 plus straight t right parenthesis equals straight u rightwards double arrow 1 plus straight t equals straight e to the power of straight u rightwards double arrow straight t equals straight e to the power of straight u space minus space 1 fraction numerator dt over denominator 1 plus straight t end fraction equals du straight t equals 0 space then space straight u equals 0 space and space straight t equals 1 space then space straight u equals log 2 straight I equals integral subscript 0 superscript log 2 end superscript space fraction numerator du over denominator square root of open parentheses straight e to the power of straight u minus 1 close parentheses open parentheses 2 plus straight e to the power of straight u minus 1 close parentheses end root end fraction straight I equals integral subscript 0 superscript log 2 end superscript space fraction numerator du over denominator square root of open parentheses straight e to the power of straight u minus 1 close parentheses open parentheses straight e to the power of straight u plus 1 close parentheses end root end fraction straight I equals integral subscript 0 superscript log 2 end superscript space fraction numerator du over denominator square root of straight e to the power of 2 straight u end exponent minus 1 end root end fraction straight I equals integral subscript 0 superscript log 2 end superscript space fraction numerator straight e to the power of negative straight u end exponent du over denominator square root of straight e to the power of negative 2 straight u end exponent straight e to the power of 2 straight u end exponent minus straight e to the power of negative 2 straight u end exponent end root end fraction straight I equals integral subscript 0 superscript log 2 end superscript fraction numerator straight e to the power of negative straight u end exponent du over denominator square root of 1 minus straight e to the power of negative 2 straight u end exponent end root end fraction Put space straight e to the power of negative straight u end exponent equals straight z minus straight e to the power of negative straight u end exponent du equals dz rightwards double arrow straight e to the power of negative straight u end exponent du equals negative dz straight u equals 0 space then space straight z equals 1 space and space straight u equals log 2 space then space straight z equals 1 half straight I equals integral subscript 1 superscript 1 half end superscript space fraction numerator negative dz over denominator square root of 1 minus straight z squared end root end fraction straight I equals integral subscript 1 half end subscript superscript 1 space fraction numerator dz over denominator square root of 1 minus straight z squared end root end fraction Integrate space it space further space using space formula space we space get space answer space as space straight pi over 3 end style

Answered by Sneha shidid | 25 Feb, 2019, 10:16: AM

CBSE Class 12-science Answered

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CBSE Class 12-science Answered

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