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.

Set
$$ I = \int \frac{\sin x}{\sin x - \cos x} dx = \int 1 + \frac{\cos x}{\sin x - \cos x} dx$$
Therefore:
$$ 2I = \int 1 + \frac{\sin x +\cos x}{\sin x - \cos x} dx $$
$$ 2I = x + \log(\sin x - \cos x) + C$$
$$ I = \frac{x}{2} + \frac{1}{2} \log(\sin x - \cos x) + C$$
