find the definite integral of
 
a) sec x/ (1+ 2sin square x)    from 0 to pi /4
 
b)  1/ (cosx into square root of blank end rootcos 2x  ) from - pi/4 to pi/4

Asked by ahuja8087 | 31st Oct, 2016, 09:10: PM

Expert Answer:

begin mathsize 16px style I equals integral subscript 0 superscript straight pi over 4 end superscript fraction numerator secx over denominator 1 plus 2 sin squared straight x end fraction dx straight I equals integral subscript 0 superscript straight pi over 4 end superscript fraction numerator cosx over denominator open parentheses cos squared straight x close parentheses open parentheses 1 plus 2 sin squared straight x close parentheses end fraction dx straight I equals integral subscript 0 superscript straight pi over 4 end superscript fraction numerator cosx over denominator negative open parentheses sin squared straight x minus 1 close parentheses open parentheses 1 plus 2 sin squared straight x close parentheses end fraction dx Substitute space straight u space equals space sinx space space space space fraction numerator space du over denominator dx end fraction equals space cosx space space space space space space when space straight x space equals space 0 space then space space straight u space equals 0 space space comma space space straight x space equals space straight pi over 4 space then space space straight u space equals fraction numerator 1 over denominator square root of 2 end fraction straight I space equals negative space integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript fraction numerator 1 over denominator open parentheses straight u squared minus 1 close parentheses open parentheses 2 straight u squared plus 1 close parentheses end fraction du straight I space equals negative space integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript minus fraction numerator 2 over denominator 3 open parentheses 2 straight u squared plus 1 close parentheses end fraction minus fraction numerator 1 over denominator 6 open parentheses straight u plus 1 close parentheses end fraction plus fraction numerator 1 over denominator 6 open parentheses straight u minus 1 close parentheses end fraction du straight I space equals space minus open square brackets space minus 2 over 3 integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript fraction numerator 1 over denominator open parentheses 2 straight u squared plus 1 close parentheses end fraction du minus 1 over 6 integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript fraction numerator 1 over denominator open parentheses straight u plus 1 close parentheses end fraction du plus 1 over 6 integral subscript 0 superscript fraction numerator 1 over denominator square root of 2 end fraction end superscript fraction numerator 1 over denominator open parentheses straight u minus 1 close parentheses end fraction du close square brackets........ left parenthesis straight i right parenthesis We space will space solve space only space integral space put space limits space at space the space end straight I subscript 1 equals integral fraction numerator 1 over denominator open parentheses 2 straight u squared plus 1 close parentheses end fraction du substitute space straight v space equals space square root of 2 space straight u space space straight I subscript 1 equals fraction numerator 1 over denominator square root of 2 end fraction integral fraction numerator 1 over denominator open parentheses straight v squared plus 1 close parentheses end fraction dv straight I subscript 1 equals fraction numerator tan to the power of negative 1 end exponent left parenthesis straight v right parenthesis over denominator square root of 2 end fraction straight I subscript 1 equals fraction numerator tan to the power of negative 1 end exponent left parenthesis square root of 2 straight u right parenthesis over denominator square root of 2 end fraction............. left parenthesis ii right parenthesis straight I subscript 2 equals integral fraction numerator 1 over denominator open parentheses straight u plus 1 close parentheses end fraction du substitute space straight v space equals space straight u space plus 1 straight I subscript 2 equals integral 1 over straight v dv straight I subscript 2 equals log open vertical bar v close vertical bar equals log open vertical bar left parenthesis straight u plus 1 right parenthesis close vertical bar.................. left parenthesis iii right parenthesis straight I subscript 2 equals integral fraction numerator 1 over denominator open parentheses straight u minus 1 close parentheses end fraction du substitute space straight v space equals space straight u space minus 1 straight I subscript 2 equals integral 1 over straight v dv straight I subscript 2 equals log open vertical bar v close vertical bar equals log open vertical bar left parenthesis straight u minus 1 right parenthesis close vertical bar.................. left parenthesis iv right parenthesis Substitute space left parenthesis ii right parenthesis comma space left parenthesis iii right parenthesis space space and space left parenthesis iv right parenthesis space space in space left parenthesis straight i right parenthesis space we space get comma straight I space equals space minus open square brackets space minus 2 over 3 cross times fraction numerator tan to the power of negative 1 end exponent left parenthesis square root of 2 straight u right parenthesis over denominator square root of 2 end fraction minus 1 over 6 log open vertical bar left parenthesis straight u plus 1 right parenthesis close vertical bar plus 1 over 6 log open vertical bar left parenthesis straight u minus 1 right parenthesis close vertical bar close square brackets to the power of fraction numerator 1 over denominator square root of 2 end fraction end exponent subscript 0 straight I equals negative open square brackets space minus 2 over 3 cross times fraction numerator tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis over denominator square root of 2 end fraction minus 1 over 6 log left parenthesis fraction numerator 1 over denominator square root of 2 end fraction plus 1 right parenthesis plus 1 over 6 log left parenthesis fraction numerator 1 over denominator square root of 2 end fraction minus 1 right parenthesis close square brackets plus open square brackets space minus 2 over 3 cross times fraction numerator tan to the power of negative 1 end exponent left parenthesis 0 right parenthesis over denominator square root of 2 end fraction minus 1 over 6 log left parenthesis 1 right parenthesis plus 1 over 6 log left parenthesis 1 right parenthesis close square brackets I space equals space minus open square brackets space minus 2 over 3 cross times fraction numerator begin display style straight pi over 4 end style over denominator square root of 2 end fraction minus 1 over 6 log left parenthesis fraction numerator 1 over denominator square root of 2 end fraction plus 1 right parenthesis plus 1 over 6 log left parenthesis fraction numerator 1 over denominator square root of 2 end fraction minus 1 right parenthesis close square brackets plus 0 I space equals 1 over 6 open square brackets fraction numerator straight pi over denominator square root of 2 end fraction minus log open parentheses fraction numerator 1 plus square root of 2 over denominator 1 minus square root of 2 end fraction close parentheses close square brackets end style

Answered by Vijaykumar Wani | 3rd Nov, 2016, 09:54: AM

Concept Videos

This video is based on evaluating definite integrals.

This video covers examples evaluating Definite Integrals.

This video is based on evaluating definite integrals.

Practice Test
All Questions Ask Doubt