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integral sinx/sinx-cosx dx=?

$\int \frac{\sin x}{\sin x - \cos x} ⅆ x = \int \frac{\sin x}{\sin x - \cos x} \times \frac{\sin x + \cos x}{\sin x + \cos x} ⅆ x = \int \frac{\sin x (\sin x + \cos x)}{\sin^{2} x - \cos^{2} x} ⅆ x = - \int \frac{\sin^{2} x + \sin x \cos x}{(\cos^{2} x - \sin^{2} x)} ⅆ x = - \frac{1}{2} \int \frac{2 (\sin^{2} x + \sin x \cos x)}{(\cos^{2} x - \sin^{2} x)} ⅆ x = - \frac{1}{2} \int \frac{2 \sin^{2} x + 2 \sin x \cos x}{\cos 2 x} ⅆ x = - \frac{1}{2} \int \frac{1 - \cos 2 x + \sin 2 x}{\cos 2 x} ⅆ x = - \frac{1}{2} \int \frac{1}{\cos 2 x} ⅆ x - \frac{1}{2} \int \frac{- \cos 2 x}{\cos 2 x} ⅆ x - \frac{1}{2} \int \frac{\sin 2 x}{\cos 2 x} ⅆ x = - \frac{1}{2} \int \sec 2 x ⅆ x + \frac{1}{2} \int 1 ⅆ x - \frac{1}{2} \int \tan 2 x ⅆ x = - \frac{1}{2} (\frac{1}{2} \ln (\sec 2 x + \tan 2 x)) + \frac{1}{2} (x) - \frac{1}{2} (\ln (\sec 2 x)) + C = \frac{1}{4} \ln (\sec 2 x + \tan 2 x) + \frac{x}{2} - \frac{1}{2} \ln (\sec 2 x) + C H e n c e, \int \frac{\sin x}{\sin x - \cos x} ⅆ x = \frac{1}{4} \ln (\sec 2 x + \tan 2 x) + \frac{x}{2} - \frac{1}{2} \ln (\sec 2 x) + C$

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