Please solve the following question

Asked by Balbir | 7th Oct, 2017, 11:29: PM

Expert Answer:

begin mathsize 16px style integral fraction numerator dx over denominator 2 plus 3 cos squared straight x end fraction

Dividing space by space cos squared straight x space to space numerator space and space denominator.
integral fraction numerator begin display style fraction numerator 1 over denominator begin display style cos squared end style begin display style straight x end style end fraction end style over denominator begin display style fraction numerator 2 over denominator begin display style cos squared end style begin display style straight x end style end fraction plus fraction numerator 3 cos squared straight x over denominator begin display style cos squared end style begin display style straight x end style end fraction end style end fraction dx
integral fraction numerator sec squared straight x over denominator 2 sec squared straight x plus 3 end fraction dx
integral fraction numerator sec squared straight x over denominator 2 left parenthesis 1 plus tan squared straight x right parenthesis plus 3 end fraction dx
Put space tanx space equals space straight t space rightwards double arrow sec squared xdx equals dt
integral fraction numerator dt over denominator 2 left parenthesis 1 plus straight t squared right parenthesis plus 3 end fraction
integral fraction numerator dt over denominator 2 plus 2 straight t squared plus 3 end fraction
integral fraction numerator dt over denominator 2 straight t squared plus 5 end fraction
solve space it space further space using space integration space formula. end style

Answered by Sneha shidid | 21st Dec, 2017, 03:18: PM