Prove that: `int_0^(2pi)(xsin^(2n)x)/(sin^(2n)+cos^(2n)x)dx`

Answer

`pi^(2)`

Explanation

Let `I=int_(0)^(2pi)(xsin^(2nx))/(sin^(2)x+cos^(2n)x)dx` . . . (i)
`I = int_(0)^(2pi)((2pi-x)[sin(2pi-x)]^(2n))/([sin (2pi-x)] ^(2n)+[cos(2pi-x)]^(2n))dx`
`[:' int_(a)^(b)f(x)dx=int_(0)^(a)f(a-x)dx]`
`I=int_(0)^(2pi)((2pi-x)*sin^(2n)x)/(sin^(2n)x+cos^(2n)x)dx`
`rArr I=int_(0)^(2pi)(2pisin^(2n)x)/(sin^(2n)x+cos^(2n)x)dx-int_(0)^(2pi)(xsin^(2n)x)/(sin^(2n)x+cos^(2n)x)dx`
`rArr I=int_(0)^(2pi)(2pisin^(2n))/(sin^(2n)x+cos^(2n)x)dx-I` [ from Eq . (i)]
`rArr I= int_(0)^(2pi)(pisin^(2n)x)/(sin^(2n)x+cos^(2n)x)dx`
`rArr I=pi[int_(0)^(pi)(pisin^(2n)x)/(sin^(2n)x+cos^(2n)x)dx +int_(0)^(pi)(sin6^(2n)(2pi-x))/(sin^(2n)(2pi-x)+cos^(2n)(2pi-x))dx]`
`[ " using property" int _(0)^(2a)f(x)dx = int_(0)^(a)[f(x)+f(2a-x)]dx]`
`I=pi [ int_(0)^(pi) (sin^(2n)xdx)/(sin^(2n)x+cos^(2n)x)dx+int_(0)^(pi)(sin^(2n)x)/(sin^(2n)x+cos^(2n)x)]`
`rArr I= 2piint_(0)^(pi)(sin^(2n)xdx)/(sin^(2n)x+cos^(2n)x)dx`
`rArr I=4pi[int_(0)^(pi//2)(sin^(2n)x dx)/(sin^(2n)x+cos^(2n)x)dx]` . . . (ii)
`rArr I=4pi int_(0)^(pi//2)(sin^(2n)(pi//2-x))/(sin^(2n)(pi//2-x)+cos^(2n)(pi//2-x))dx`
`rArr I= 4pi int_(0)^(pi//2)(cos^(2n)x)/(cos^(2n)x+sin^(2n)x)dx` . . . (iii)
On adding Eqs . (ii) and (iii) , we get
`2I=4piint_(0)^(pi//2)(sin^(2n)x+cos^(2n)x)/(sin^(2n)x+cos^(2n)x)dx`
`rArr2I=4pi int_(0)^(pi//2)1 dx=4pi [x] _(0)^(pi//2)=4pi *(pi)/(2)`
`rArrI=pi^(2)`