`int(3+2cosx)/((2+3cosx)^(2))dx` is equal toA. `((sinx)/(2+3cosx))+C`B. `((2cosx)/(2+3sinx))+C`C. `((2cosx)/(2+3cosx))+C`D. `((2sinx)/(2+3sinx))+C`

`int(3+2cosx)/((2+3cosx)^(2))dx` is equal toA. `((sinx)/(2+3cosx))+C`B

1.	`int(3+2cosx)/((2+3cosx)^(2))dx` is equal toA. `((sinx)/(2+3cosx))+C`B. `((2cosx)/(2+3sinx))+C`C. `((2cosx)/(2+3cosx))+C`D. `((2sinx)/(2+3sinx))+C`
Answer» Correct Answer - A Divide numerator and denominator by `sin^(2)x`, then `int(("3 cosec"^(2)x+"2 cosec x cot x"))/(("2 cosec x "+3 cot x)^(2))dx` Put `" 2 cosec x "+"3 cot x = t` ` therefore" "(-"2 cosec x cot x "-"3 cosec"^(2)x)dx=dt` `=int-(dt)/(t^(2))=(1)/(t)+C=(1)/("2 cosec x"+"3 cot x")+C` `=(sinx)/((2+3 cos x))+C`