$\sin(nx) = \sin x \cos(n-1)x + \cos x \sin(n-1)x$
$\\$ $\int\frac{\sin x \cos((n-1)x) + \cos x \sin((n-1)x)}{\sin x} \text{d}x $
or:
$ \int (\cos((n-1)x) + \frac{\cos x \sin((n-1)x)}{\sin x})\text{d}x $
Which equals to:
$\frac{\sin((n-1)x)}{n-1}+\int\frac{\cos x \sin((n-1)x)}{\sin x}\text{d}x$
So, the problem is evaluating,
$\int\frac{\cos x \sin((n-1)x)}{\sin x})\text{d}x$
From some trig identities, we will get that,
$\cos x \sin((n-1)x) = \frac{1}{2}(\sin(nx) + \sin((n-2)x)$
Hence our "problematic" integral becomes to:
$ \int\frac{\frac{1}{2}(\sin(nx) + \sin((n-2)x)}{\sin x} $
Or:
$ \frac{1}{2}\int\frac{\sin(nx)}{\sin x}+\frac{1}{2}\int\frac{\sin(n-2)x}{\sin x}$
And if we go back and say note our original integral as:
$ I_n=\int\frac{\sin(nx)}{\sin x}\text{d}x $
by our above observation, we can conclude:
$ I_n=\frac{\sin((n-1)x}{n-1}+\frac{1}{2}I_n+\frac{1}{2}I_{n-2}$
or:
$ I_n=2 \frac{\sin((n-1)x)}{n-1}+I_{n-2}$
Could you proceed?