Answer
$-\ln (1-\sin x)+C$
Work Step by Step
$\int\frac{\cos x}{1-\sin x}dx$
Let $u=1-\sin x$. Then $du=-\cos x dx$, and $\cos xdx=-du$.
$=\int\frac{-du}{u}$
$=-\ln u+C$
$=-\ln (1-\sin x)+C$
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