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$\int \frac{d x}{1-\sin x}=$
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The correct answer is:
$\sec x+\tan x+c$
$\begin{aligned} & \int \frac{d x}{1-\sin x}=\int \frac{(1+\sin x)}{1-\sin ^2 x} d x \\ & =\int \sec ^2 x d x+\int \tan x \cdot \sec x d x=\tan x+\sec x+c .\end{aligned}$
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