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

Explanation

`I=int(1)/(3+sin2x)dx`
`=int(1)/(3(sin^(2)x+cos^(2)x)+2sinx cosx)dx`
`=int(sec^(2)x)/(3tan^(2)x+2tanx+3)dx`
`" "[ " Dividing " N^(r) " and " D^(r) "by " cos^(2)x]`
Putting `tanx=t` and `sec^(2)xdx=dt, ` we get
`I=int(dt)/(3t^(2)+2t+3)=(1)/(3)int(dt)/(t^(2)+(2)/(3)t+1)`
`=(1)/(3)int(dt)/((t+(1)/(3))^(2)+((2sqrt(2))/(3))^(2))`
`:. I=(1)/(3)(1)/(((2sqrt(2))/(3)))tan^(-1)((t+(1)/(3))/((2sqrt(2))/(3)))+C`
`=(1)/(2sqrt(2))tan^(-1)((3tanx+1)/(2sqrt(2)))+C`