`int_(pi//4)^(pi//3)(1+sec\ x)^2dx`

Now

`(1+sec\ x)^2` `=1+2\ sec\ x+sec^2x`

So

`int_(pi//4)^(pi//3)(1+sec\ x)^2dx =int_(pi//4)^(pi//3)(1+2\ sec\ x+sec^2 x)dx`

`=[x+2\ ln\ |sec\ x+tan\ x|+ ` `{:tan\ x ]_(pi//4)^(pi//3)`

`=[(pi/3+2(1.31696)+1.7321)-` `{:(pi/4+2(0.88137)+1)]`

`=1.8651`