Evaluate: `int1/(3+sin2x)dx`

Answer

`(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c`

Explanation

`I=int(dx)/(3+sin2x)`
`=int(dx)/(3+sinx cosx)`
` =int(sec^(2)xdx)/(3sec^(2)x+2tanx)" "("Dividing Nr. and Dr. by " cos^(2)x)`
` =int(sec^(2)xdx)/(3tan^(2)x+2tanx+3) `
` =int(dt)/(3t^(2)+2t+3)" " ("Putting " tanx=t)`
`=(1)/(3) int (dt)/(t^(2)+(2)/(3)t+1)`
` = (1)/(3) int (dt)/((t+(1)/(3))^(2)+(8)/(9))`
` = (1)/(3)* (1)/(sqrt(8)//3) "tan"^(-1) ((t+(1)/(3)))/((sqrt(8))/(3))+c `
` =(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c`