Evaluate: `int_(pi//4)^(pi//4)(x+pi//4)/(2-cos2x)dx`

Explanation

`I=int_(-(pi)/(4))^((pi)/(4))(x+(pi)/(4))/(2-cos2x)dx=int_(-(pi)/(4))^((pi)/(4))(x)/(2-cos2x)dx+(pi)/(4)int_(-(pi)/(4))^((pi)/(4))(dx)/(2-cos2x)`
`0+(pi)/(4)*2int_(0)^((pi)/(4))(dx)/(2-cos2x)`
(As the first function is odd and second even)
`=(pi)/(2)*int_(0)^((pi)/(4))(dx)/(2-cos2x)`
`=(pi)/(2)*int_(0)^((pi)/(4))(dx)/(2-(2cos^(2)x-1))`
`=(pi)/(2)*int_(0)^((pi)/(4)(dx)/(2sin^(2)x+1)` ltbr `=(pi)/(2)*int_(0)^((pi)/(4))(sec^(2)x)/(3tan^(2)x+1)dx`
Let `t=sqrt(3)tanx`, so that `dt=sqrt(3)sec^(2)xdx`. Also limits `0` and `(pi)/(4)` are changed to `0` and `sqrt(3)`.
The integral becomes
`(pi)/(2)*(1)/(sqrt(3))int_(0)^(sqrt(3))(dt)/(t^(2)+1)=(pi)/(2sqrt(3))[tan^(-1)t]_(0)^(sqrt(3))`
`=(pi)/(2sqrt(3))*tan^(-1)sqrt(3)=(pi)/(2sqrt(3)xx(pi)/(3)`
`=(pi^(2))/(6sqrt(3))`