Prove that begin mathsize 20px style integral fraction numerator 1 over denominator 9 plus 16 cos squared x end fraction d x equals 1 over 15 tan to the power of negative 1 end exponent open parentheses 5 over 3 tan x close parentheses end style

Asked by sunil2791 | 11th Jun, 2017, 06:15: AM

Expert Answer:

begin mathsize 16px style integral fraction numerator 1 over denominator 9 plus 16 cos squared straight x end fraction dx
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 dx end style over denominator begin display style fraction numerator 9 over denominator cos squared straight x end fraction plus fraction numerator 16 cos squared straight x over denominator cos squared straight x end fraction end style end fraction
integral fraction numerator sec squared xdx over denominator 9 sec squared straight x plus 16 end fraction
integral fraction numerator begin display style sec squared xdx end style over denominator begin display style 9 open parentheses 1 plus tan squared straight x close parentheses plus 16 end style end fraction
Put space tanx equals straight t
sec squared xdx equals dt
integral fraction numerator begin display style dt end style over denominator begin display style 9 open parentheses 1 plus straight t squared close parentheses plus 16 end style end fraction
Use space formula space integral fraction numerator 1 over denominator straight x squared plus straight a squared end fraction dx


end style

Answered by Sneha shidid | 11th Jun, 2017, 01:12: PM