Please solve the following

Asked by Balbir | 6th Oct, 2017, 12:11: PM

Expert Answer:

begin mathsize 16px style integral fraction numerator dx over denominator 2 cos squared straight x space sinx space cosx end fraction
integral fraction numerator sec squared xdx over denominator 2 sinxcosx end fraction
integral fraction numerator sec squared xdx over denominator sin begin display style 2 end style begin display style straight x end style end fraction
integral fraction numerator begin display style sec squared xdx end style over denominator begin display style fraction numerator 2 tanx over denominator 1 plus tan squared straight x end fraction end style end fraction
integral fraction numerator sec squared xdx over denominator 2 tanx end fraction cross times open parentheses 1 plus tan squared straight x close parentheses
Put space tan space straight x space equals space straight t space rightwards double arrow space sec squared xdx space equals space dt
integral fraction numerator 1 plus straight t squared over denominator 2 straight t end fraction dt
integral open parentheses fraction numerator 1 over denominator 2 straight t end fraction plus straight t over 2 close parentheses dt space
1 half log space open vertical bar straight t close vertical bar space plus space straight t squared over 4 space plus space straight c
fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction log space open vertical bar tan straight x close vertical bar space plus space fraction numerator begin display style tan squared straight x end style over denominator begin display style 4 end style end fraction space plus space straight c

end style

Answered by Sneha shidid | 22nd Dec, 2017, 09:45: AM