if n is a +ive Z, and..... (Calculus)

Question

Sorry wasn't able to type the question.... also have proved it, just need help deducing it

Answer 1

The first part is covered by Proof by Induction, and gives us the iteration formula
I_(n-1) – I_n = 2π

For brevity and clarity I am going to use the notation int[f(x)] to represent the integral, from 0 to 2π, of f(x) with respect to x.

To prove: int[sin^2(nx/2) / sin^2(x/2)] = 2nπ

sin^2(nx/2) / sin^2(x/2) = 2(1 – cos(nx)) / ({2(1 – cos(x))} = (1 – cos(nx)) / (1 – cos(x))
Therefore,
int[sin^2(nx/2) / sin^2(x/2)] = int[(1 – cos(nx)) / (1 – cos(x))]
int[] = int[(1) / (1 – cos(x))] –  int[(cos(nx)) / (1 – cos(x))]
But, from part 1,  int[(cos(nx)) / (1 – cos(x))] = I_n, therefore,
int[] = int[(1) / (1 – cos(x))] – I_n
int[] = int[(1) / (1 – cos(x))] – I_(n-1) + 2π
int[] = int[(1) / (1 – cos(x))] – int[(cos((n-1)x)) / (1 – cos(x))]  + 2π
int[] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π, i.e.
int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π
The above is an iteration formula, with the nth term on the lhs and the (n-1)th term on the rhs.
We can expand this iteration sequence as follows,
int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-2)x)) / (1 – cos(x))]  + 2(2π)
int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-3)x)) / (1 – cos(x))]  + 3(2π)
...
int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-(n-1))x)) / (1 – cos(x))]  + (n-1)(2π)
int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos(x)) / (1 – cos(x))]  + (n-1)(2π)
int[(1 – cos(nx)) / (1 – cos(x))] = int[ (1) ] + (n-1)(2π)
int[(1 – cos(nx)) / (1 – cos(x))] = 2π + (n-1)(2π)
int[(1 – cos(nx)) / (1 – cos(x))] = 2nπ
int[sin^2(nx/2) / sin^2(x/2)] = 2nπ