$\int \frac{\sin x}{\sin x-\cos x} d x$ is equal to :

$\frac{x}{2}+\frac{1}{2} \log (\sin x-\cos x)+c$

Let sin x = A( sin x – cos x) + B. d.c of (sin x – cos x)

or sin x = A ( sin x – cos x) + B ( cos x + sin x)

or sin x= (A + B ) sin x + (B –A) cos x

equating the coefficient of sin x and cos x, we get

A + B = 1 and B – A = 0

A = 1/2, B = 1/2

$I =\int \frac{\frac{1}{2}(\sin x-\cos x)+\frac{1}{2}(\cos x+\sin x)}{\sin x-\cos x}$

$=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\sin x-\cos x} d x=\frac{x}{2}+\frac{1}{2} \log (\sin x-\cos x)+c$

Hence (2) is the correct answer.