Evaluate `int_(0)^(2pi) (dx)/(1+3cos^(2)x)`

Explanation

`I=int_(0)^(2pi)(dx)/(1+3cos^(2)x)`
`=int_(0)^(2pi)(sec^(2) xdx)/(sec^(2)x+3)`
`=int_(0)^(2pi)(sec^(2) xdx)/(4+tan^(2)x)`
Here we cannot substitute `tan x=t` as `tanx` is discontinuous in `(0,2pi)`
Let `f(x)=(sec^(2)x)/(4+tan^(2)x)`
`:f(2pix)=f(x)`
`:.I=2 int_(0)^(pi)(sec^(2) xdx)/(4+tan^(2)x)`
Also `f(pi-x)=f(x)`
`:. I=4 int_(0)^(pi//2)(sec^(2) xdx)/(4+tan^(2)x`
`=4 int_(0)^(oo) (dt)/(4+t^(2))` (Putting `tanx=t` as `tanx` is continuous in `(0,pi//2)`
`=4/2[tan^(-1)t/2]_(0)^(oo)`
`=2[ta ^(-1)oo-tan^(-1)0]`
`=pi`