Show that the partial sum $$f_N(x)=\frac{4}{π}\sum^N_{n=1}\frac{\sin((2n-1)x)}{2n-1}$$ may be written as $$f_N(x)=\frac{2}{π}\int_0^x\frac{\sin(2Nt)}{\sin(t)}\,dt$$
The original question is 'Sketch the odd extension for the function $f(x)=1,\ 0<x<π .$ and find its Fourier sine series expansion.' (and I found the above)
First I need to take the derivative of $f_N(x)$ which is $$\frac{4}{π}\sum_{n=1}^N\cos((2n-1)x)$$
but what should I do next? If anyone could help me out that would be great!
$$\frac{df_N(x)}{dx}=\frac{4}{π}\sum_{n=1}^N\cos((2n-1)x)$$
Using $$ \sin(2Nx)= 2 \sin(x)\sum_{n=1}^N\cos((2n-1)x)$$ The proof by induction being:
$N=1$ $$LHS: \sin(2x)=2\sin(x)\cos(x)$$ $$RHS: 2\sin(x)\sum_{n=1}^N\cos((2n-1)x)=2\sin(x)\cos(x)$$
Assume true for N-1: $$\sin2((N-1)x)=2\sin(x)\sum_{n=1}^{N-1}\cos((2n-1)x)$$
Using $2\sin(a)\cos(b)= \sin(a+b)+\sin(a-b)$ $$2\sin(x)\cos((2n-1)x)=\sin(2nx)-\sin((2n-2)x)$$
so: $$\sin(2Nx)=2\sin(x)\cos((2N-1)x)+\sin(2x(N-1))$$ $$\sin(2Nx)=2\sin(x)\cos((2N-1)x)+2\sin(x)\sum_{n=1}^{N-1}\cos((2n-1)x)$$ $$\sin(2Nx)=2\sin(x)\sum_{n=1}^{N}\cos((2n-1)x)$$ $$\frac{1}{2}\sin(2Nx)=\sin(x)\sum_{n=1}^{N}\cos((2n-1)x)$$ $$\frac{1}{2}\sin(2Nx)=\sin(x)\frac{df_N(x)}{dx}$$
So $$\frac{df_N(x)}{dx}=\frac{4}{2\pi}\frac{\sin(2Nx)}{\sin(x)}$$ $$f_N(x)=\frac{2}{\pi}\int_0^x\frac{\sin(2Nt)}{\sin(t)}dt$$