Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.

Let $$I_{1} = \int \frac{\sin x}{\sin x - \cos x}dx, \quad I_{2} = \int \frac{\cos x}{\sin x - \cos x}dx$$ 
Then $$I_{1}-I_{2} = \int 1\, dx =  x + c_{1}$$
$$I_{1}+I_{2} = \int \frac{\sin x + \cos x}{\sin x - \cos x} dx = \ln(\sin x - \cos x) + c_{2} $$ 
Then solve simultaneously 
$$I_{1} = \frac{1}{2}\left(x+ \ln(\sin x - \cos x) \right) + c$$
