Functions `f,g:R→R` are defined, respectively, by `f(x)=x^2+3x+1,g(x)=2x−3`, find `fof`

Explanation

Let `I=int_(-pi//3)^(pi//3) (pi+4x^(3))/(2-cos(|x|+(pi)/3))dx+int_(-pi//3)^(pi/3)(4x^(3))/(2-cos(|x|+(pi)/3))dx`
The second integral becomes zero as integrand being an odd function of `x`.
`:. I=2pi int_(0)^(pi//3)(dx)/(2-cos(x+(pi)/3))`
Let `x+pi//3=y` or `dx=dy`.
Also as `xto0,y to pi//3`, and as `x to pi//3, yto2pi//3`.
`:. I=2pi int_(pi//3)^(2pi//3)(dy)/(2-cosy)`
`=2pi int_(pi//3)^(2pi//3) (dy)/(2-(1-tan^(2)y//2)/(1+tan^(2)y//2))`
`=2pi int_(pi//3)^(2pi//3)(1/2sec^(2)y//2)/(tan^(2)y//2+(1//sqrt(3))^(2))dy`
`=(4pisqrt(3))/3[tan^(-1)(sqrt(3) tan y//2)]_(pi//3)^(2pi//3)`
`=(4pi)/(sqrt(3))[tan^(-1)3-tan^(-1)1]`
`=(4pi)/(sqrt(3))[tan^(-1)3-pi//4]`