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CBSE Class 12-science Answered

 
Evaluate
 
a)integral space 1 divided by left parenthesis space cos cubed space x space square root of sin 2 x end root right parenthesis space d x space f r o m space 0 space t o space pi divided by 4
b) integral subscript negative pi divided by 2 end subscript superscript pi divided by 2 end superscript space cos x space divided by left parenthesis space 1 space plus space e hat x right parenthesis space d x
 
Asked by ahuja8087 | 01 Nov, 2016, 04:32: PM
answered-by-expert Expert Answer
begin mathsize 16px style integral subscript 0 superscript straight pi divided by 4 end superscript fraction numerator 1 over denominator cos cubed straight x square root of sin 2 straight x end root end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript fraction numerator 1 over denominator cos cubed straight x square root of 2 sinx space cosx end root end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript sec squared straight x fraction numerator 1 over denominator cosx square root of 2 sinx space cosx end root end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript sec squared straight x fraction numerator 1 over denominator begin display style fraction numerator cos squared straight x over denominator cosx end fraction square root of 2 sinx space cosx end root end style end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript sec squared straight x fraction numerator 1 over denominator begin display style cos squared straight x square root of fraction numerator 2 sinx space cosx over denominator cos squared straight x end fraction end root end style end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript sec squared straight x fraction numerator sec squared straight x over denominator begin display style square root of 2 space tanx end root end style end fraction dx equals integral subscript 0 superscript straight pi divided by 4 end superscript sec squared straight x fraction numerator open parentheses tan squared straight x plus 1 close parentheses over denominator begin display style square root of 2 space tanx end root end style end fraction dx Put space straight u equals tanx rightwards double arrow du over dx equals sec squared straight x. space When space straight x equals 0 comma space straight u equals 0 space and space when space straight x equals straight pi over 4 comma space straight u equals 1 So comma space integral subscript 0 superscript 1 sec squared straight x fraction numerator open parentheses tan squared straight x plus 1 close parentheses over denominator begin display style square root of 2 square root of tanx end style end fraction dx equals integral subscript 0 superscript 1 fraction numerator open parentheses straight u squared plus 1 close parentheses over denominator begin display style square root of 2 square root of straight u end style end fraction du Now space put space straight v equals square root of straight u rightwards double arrow dv over du equals fraction numerator 1 over denominator 2 square root of straight u end fraction rightwards double arrow dv equals fraction numerator 1 over denominator 2 square root of straight u end fraction du. space When space straight u equals 0 comma space straight v equals 0 space and space when space straight u equals 1 comma space straight v equals 1 So comma space integral subscript 0 superscript 1 space space end superscript fraction numerator open parentheses straight u squared plus 1 close parentheses over denominator begin display style square root of 2 square root of straight u end style end fraction du equals integral subscript 0 superscript 1 space space end superscript space space left parenthesis straight v to the power of 4 plus 1 right parenthesis space dv equals open curly brackets open square brackets straight v to the power of 5 over 5 close square brackets subscript 0 to the power of 1 space space space plus open square brackets straight v close square brackets subscript 0 to the power of 1 close curly brackets equals open square brackets open parentheses 1 fifth minus 0 close parentheses plus open parentheses 1 minus 0 close parentheses close square brackets equals 1 fifth plus 1 equals 6 over 5 Kindly space ask space only space 1 space query space per space question. end style
Answered by Rebecca Fernandes | 02 Nov, 2016, 12:31: PM

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