Please solve the following question

Asked by Balbir | 7th Oct, 2017, 11:16: PM

Expert Answer:

      begin mathsize 16px style integral 1 over straight x square root of fraction numerator 1 plus square root of straight x over denominator 1 minus square root of straight x end fraction end root dx
integral 1 over straight x square root of fraction numerator 1 plus square root of straight x over denominator 1 minus square root of straight x end fraction cross times fraction numerator 1 plus square root of straight x over denominator 1 plus square root of straight x end fraction end root dx
integral 1 over straight x cross times fraction numerator 1 plus square root of straight x over denominator square root of 1 minus straight x end root end fraction dx
Put space straight x space equals space sin squared straight theta rightwards double arrow square root of straight x space equals space sinθ
dx space equals space 2 sinθ space cosθdθ
integral fraction numerator 1 over denominator space sin squared straight theta end fraction cross times fraction numerator 1 plus sinθ over denominator square root of 1 minus space sin squared straight theta end root end fraction cross times 2 sinθ space cosθdθ
integral fraction numerator begin display style 1 end style over denominator begin display style space sin squared straight theta end style end fraction cross times fraction numerator begin display style 1 plus sinθ end style over denominator begin display style cos straight theta end style end fraction cross times 2 sinθ space cosθdθ
2 integral fraction numerator 1 plus sinθ over denominator sinθ end fraction dθ
2 integral open parentheses 1 over sinθ plus 1 close parentheses dθ
2 integral open parentheses cosecθ space plus space 1 close parentheses dθ
Integrate space it space using space formulae.
end style

Answered by Sneha shidid | 18th Dec, 2017, 11:10: AM