Evaluate : integral subscript 0 superscript pi space fraction numerator space x space d x over denominator 1 plus space e sin x end fraction      ,    e squared space less than space 1 space
 
Answer is  fraction numerator pi space cos to the power of negative 1 end exponent space x over denominator square root of 1 minus e squared end root end fraction

Asked by ahuja8087 | 15th Mar, 2017, 04:54: PM

Expert Answer:

begin mathsize 16px style straight I equals integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus esinx end fraction dx space space space space space space space.............. left parenthesis straight i right parenthesis
straight I equals integral subscript 0 superscript straight pi fraction numerator straight pi minus straight x over denominator 1 plus esin open parentheses straight pi minus straight x close parentheses end fraction dx............ left parenthesis ii right parenthesis
Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma
2 straight I equals integral subscript 0 superscript straight pi fraction numerator straight pi over denominator 1 plus esinx end fraction dx
straight I equals straight pi over 2 integral subscript 0 superscript straight pi fraction numerator 1 over denominator 1 plus esinx end fraction dx
space straight I equals straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus esinx end fraction dx
straight I equals straight pi integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator 1 over denominator 1 plus esin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx
straight I equals straight pi integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator 1 over denominator 1 plus ecosx end fraction dx
Use space integral fraction numerator dx over denominator straight a plus bcosx end fraction equals fraction numerator 1 over denominator square root of straight a squared minus straight b squared end root end fraction cos to the power of negative 1 end exponent open parentheses fraction numerator straight b plus acosx over denominator straight a plus bcosx end fraction close parentheses space comma space where space straight a squared greater than straight b squared
Here comma space straight a equals 1 space and space straight b equals straight e space also space straight e squared less than 1
equals fraction numerator straight pi over denominator square root of 1 minus straight e squared end root end fraction open square brackets cos to the power of negative 1 end exponent straight e minus cos to the power of negative 1 end exponent open parentheses fraction numerator straight e plus 1 over denominator 1 plus straight e end fraction close parentheses close square brackets
equals fraction numerator straight pi over denominator square root of 1 minus straight e squared end root end fraction open square brackets cos to the power of negative 1 end exponent straight e minus cos to the power of negative 1 end exponent open parentheses 1 close parentheses close square brackets
equals fraction numerator straight pi over denominator square root of 1 minus straight e squared end root end fraction open square brackets cos to the power of negative 1 end exponent straight e minus 0 close square brackets
equals fraction numerator straight pi cos to the power of negative 1 end exponent straight e over denominator square root of 1 minus straight e squared end root end fraction

end style
Dear student, the answer is as shown above. Kindly check. Since this is a definite integral so it should have a constant answer rather than that in terms of x.

Answered by Rebecca Fernandes | 16th Mar, 2017, 02:21: PM