Evaluate: `int(dx)/(9+16sin^2x)`

Answer

`(1)/(15) "tan"^(-1) (5tanx)/(3)+C`

Explanation

`int(dx)/(9+16sin^(2)x) = int (dx)/(9cos^(2)x+25sin^(2)x) =int(sec^(2)x dx)/(9+25tan^(2)x)`
` =int(dx)/(9+25z^(2)) =(1)/(25) int (dz)/(z^(2)+((3)/(5))^(2)) " " (z= tanx) `
` =(1)/(25)xx (1)/(3//5) "tan"^(-1) (z)/(3//5)+C `
`=(1)/(15) "tan"^(-1)(5z)/(3)+c =(1)/(15) "tan"^(-1) (5tanx)/(3)+C`