$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx$ by elementary methods What is an elementary way to show that for positive integer $n$
$$
\int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k}
$$
This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{d}x = n\pi$ at a beginning calculus level, that is, without using contour integration or Fourier integrals or Parseval's theorem (but with clever tricks allowed).  There, it would suffice to know that the integral from zero to $\pi$ is $\pi$.
I've tried to do this by induction but got nowhere (except that the base for $n=1$ is easy).
I've tried expanding as a series and matching terms but it got pretty messy.

Work backwards:
Let $$F(x)=x + \frac{\sin (nx)}{n} + 2 \sum_{k-1}^{n-1}\frac{\sin(kx)}{k}$$
Then 
$$F'(x)=1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
Therefore, you need to prove that 
$$\frac{\sin(nx) \sin x}{1-\cos x} = 1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx)$$
or equivalently that 
$$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)(**)$$
If there is no typo in your problem, this should be a relatively easy induction question. 
Or better, if your students know complex numbers, just do the standard 
$$z =\cos(x)+i\sin(x)$$
and calculate
$$1+2z+2z^2+..+2z^{n-1}+z^n=2(1+z+z^2+..+z^{n-1}+z^n)-1-z^n=2\frac{1-z^{n+1}}{1-z}-1-z^n \\
=\frac{2-z^{n+1}-z^n}{1-z}-1=\frac{(2-z^{n+1}-z^n)(1-\bar{z})}{(1-\cos(x))^2+\sin(x)^2}-1\\
=\frac{2-z^{n+1}-z^n-2\bar{z}+z^n+z^{n-1}}{2-2\cos(x)}-1\\
=\frac{2-z^{n+1}-2\bar{z}+z^{n-1}}{2-2\cos(x)}-1\\$$
By taking the real parts we get:
$$
\left(1 + \cos (nx) + 2 \sum_{k-1}^{n-1}\cos(kx) \right)=\frac{2-\cos((n+1)x)-2\cos(x)+\cos((n-1)x)-2+2\cos(x)}{2-2\cos(x)} \\
=\frac{-\cos((n+1)x)+\cos((n-1)x)}{2-2\cos(x)}=\frac{2 \sin(nx)\sin(x)}{2-2\cos(x)}
$$
which is exactly what you need to prove.
P.S. You can also try to prove $(**)$ by writing
$$(1-\cos(x)) \left( \sum_{k-1}^{n-1}\cos(kx) \right)=2\sin^2(\frac{x}{2})\left( \sum_{k-1}^{n-1}\cos(kx) \right)$$ and use the fact that 
$$\sum_{k-1}^{n-1}\cos(kx) \sin(\frac{x}{2})$$
is telescopic.
