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10th of March 2017
Mathematics
Q:

Evaluateintegral fraction numerator d x over denominator sin x left parenthesis 5 minus 4 cos x right parenthesis end fraction

Please also tell whether we can evaluateintegral subscript 0 superscript pi fraction numerator x d x over denominator 1 plus e sin x end fraction. if this cannot be evaluated , please explain the reason for my understanding . Thanks

Ahuja8087 Ahuja - CBSE - Class XII Science

Friday, March 10, 2017 at 18:24:PM

A:

Dear student, this is the solution to your query. 

begin mathsize 16px style Let space straight I equals integral fraction numerator dx over denominator sinx left parenthesis 5 minus 4 cosx right parenthesis end fraction
Using space sin 2 straight x equals 2 sinxcosx space and space cos 2 straight x equals cos squared straight x minus sin squared straight x
rightwards double arrow space straight I equals integral fraction numerator dx over denominator 2 sin begin display style straight x over 2 end style cos straight x over 2 open parentheses 5 minus 4 open parentheses cos squared straight x over 2 minus sin squared straight x over 2 close parentheses close parentheses end fraction
Divide space by space numerator space and space denominator space by space cos squared straight x over 2
rightwards double arrow space straight I equals integral fraction numerator begin display style fraction numerator 1 over denominator cos squared straight x over 2 end fraction end style over denominator 2 space tan straight x over 2 open parentheses 5 minus 4 open parentheses cos squared straight x over 2 minus sin squared straight x over 2 close parentheses close parentheses end fraction dx
space rightwards double arrow space straight I equals integral fraction numerator begin display style sec to the power of 4 straight x over 2 end style over denominator 2 space tan straight x over 2 open parentheses 5 open parentheses 1 plus tan squared straight x over 2 close parentheses minus 4 open parentheses 1 minus tan squared straight x over 2 close parentheses close parentheses end fraction dx
space rightwards double arrow space straight I equals integral fraction numerator begin display style sec squared straight x over 2. space sec squared straight x over 2 end style over denominator 2 space tan straight x over 2 open parentheses 5 plus 5 tan squared straight x over 2 minus 4 plus 4 tan squared straight x over 2 close parentheses end fraction dx
space rightwards double arrow space straight I equals integral fraction numerator begin display style sec squared straight x over 2. space open parentheses 1 plus tan squared straight x over 2 close parentheses end style over denominator 2 space tan straight x over 2 open parentheses 1 plus 9 space tan squared straight x over 2 close parentheses end fraction dx
Put space tan straight x over 2 equals straight t rightwards double arrow 1 half sec squared straight x over 2 space dx equals dt
rightwards double arrow space straight I equals integral fraction numerator begin display style 2 space open parentheses 1 plus straight t squared close parentheses end style over denominator 2 space straight t open parentheses 1 plus 9 space straight t squared close parentheses end fraction dt
rightwards double arrow space straight I equals integral fraction numerator begin display style 1 plus straight t squared end style over denominator straight t open parentheses 1 plus 9 space straight t squared close parentheses end fraction dt
rightwards double arrow space straight I equals integral fraction numerator begin display style 1 plus 9 straight t squared minus 8 straight t squared end style over denominator straight t open parentheses 1 plus 9 space straight t squared close parentheses end fraction dt
rightwards double arrow space straight I equals integral fraction numerator begin display style 1 end style over denominator straight t end fraction minus fraction numerator begin display style 8 straight t end style over denominator open parentheses 1 plus 9 space straight t squared close parentheses end fraction dt
rightwards double arrow space straight I equals ln vertical line straight t vertical line minus 4 over 9 ln vertical line 1 plus 9 space straight t squared vertical line
rightwards double arrow space straight I equals ln vertical line tan straight x over 2 vertical line minus 4 over 9 ln vertical line 1 plus 9 space tan squared straight x over 2 vertical line plus straight C end style

As for the pending query, we will have our expert who will solve it on tuesday or wednesday and send it to you with your other questions that you post. We will surely update you on that student. As of now, we will have to keep that query pending. Thank you student.

Friday, March 10, 2017 at 21:51:PM