Evaluate:

$I = \int \frac{\sin x \, dx}{\cos x(1 - \sin x)} = \int \frac{\sin x \cos x \, dx}{\cos^2 x(1 - \sin x)} = \int \frac{\sin x \cos x \, dx}{(1 - \sin^2 x)(1 - \sin x)} = \int \frac{\sin x \cos x \, dx}{(1 - \sin x)^2 (1 + \sin x)}$. Let $u = \sin x$, then $du = \cos x \, dx$. $I = \int \frac{u \, du}{(1 - u)^2 (1 + u)}$. Using partial fractions: $\frac{u}{(1 - u)^2 (1 + u)} = -\frac{1/4}{1 - u} + \frac{1/2}{(1 - u)^2} - \frac{1/4}{1 + u}$. Integrating each term: $I = \frac{1}{4} \ln|1 - u| + \frac{1}{2(1 - u)} - \frac{1}{4} \ln|1 + u| + C = \frac{1}{4} \ln\left|\frac{1 - \sin x}{1 + \sin x}\right| + \frac{1}{2(1 - \sin x)} + C$.