Evaluate `int_(0)^(2pi)(xcos^(2n)x)/(cos^(2n)x+sin^(2n)x)dx`

Explanation

`I=int_(0)^(2pi)(xcos^(2n)x)/(cos^(2n)x+sin^(2n)x) dx`………………1
`:. I=int_(0)^(2p)((2pi-x)cos^(2n)x)/(cos^(2n)x+sin^(2n)x)dx`………………….2
`["Using"int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
Adding 1 and 2 we get
`2I=2pi int_(0)^(2pi)(cos^(2n)x)/(ubrace(cos^(2n)x+sin^(2n)x)_(f(x)) dx`
`=4pi int_(0)^(pi)(cos^(2n)x)/(cos^(2n)x+sin^(2n)x)dx [ :' f(2pi-x)=f(x)]`
`:.2I=8pi int_(0)^(pi//2)(cos(2n)x)/(cos^(2n)x+sin^(2n)x) dx` ................3 `[ :' f(pi-x)=f(x)]`
`=8pi int_(0)^(pi//2)(cos^(2n)((pi)/2-x))/(cos^(2n)((pi)/2-x)+sin^(2n)((pi)/2-x)) dx`
[Using `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`]
`:.2I=8pi int_(0)^(pi//2)(sin^(2n)x)/(sin^(2n)x+cos^(2n)n) dx`...............4
Adding 3 and 4 we have `4I=8pi int_(0)^(pi//2) 1dx`
`:.I=pi^(2)`