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Please give me the complete solution of 

Asked by Balbir 23rd October 2017, 8:46 PM
Answered by Expert
Answer:
begin mathsize 16px style integral fraction numerator dx over denominator straight x square root of ax minus straight x squared end root end fraction
integral fraction numerator dx over denominator straight x square root of begin display style straight a squared over 4 end style minus begin display style straight a squared over 4 end style plus ax minus straight x squared end root end fraction
integral fraction numerator dx over denominator straight x square root of open parentheses begin display style straight a over 2 end style close parentheses squared minus open parentheses straight x squared minus ax plus straight a squared over 4 close parentheses end root end fraction
integral fraction numerator dx over denominator straight x square root of open parentheses straight a over 2 close parentheses squared minus open parentheses straight x minus begin display style straight a over 2 end style close parentheses squared end root end fraction
integral fraction numerator dx over denominator begin display style straight x over 2 end style square root of straight a squared minus open parentheses 2 straight x minus straight a close parentheses squared end root end fraction
2 straight x minus straight a equals straight t space rightwards double arrow space 2 dx space equals space dt space rightwards double arrow space dx space equals space dt over 2
2 straight x space equals space straight t plus straight a
straight x space equals space fraction numerator straight t plus straight a over denominator 2 end fraction
1 over straight x equals fraction numerator 2 over denominator straight t plus straight a end fraction
Using space this space substitution space and space further space using space trigonometric
substitution space integrate space further.
end style
Answered by Expert 14th December 2017, 4:52 PM
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