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$\int \frac{d x}{\cos 2 x-\cos ^{2} x}=$
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The correct answer is:
$\cot x+\mathrm{c}$
(B)
$\begin{aligned} \int \frac{d x}{\cos 2 x-\cos ^{2} x} &=\int \frac{1}{2 \cos ^{2} x-1-\cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x-1} d x \\ &=\int \frac{-1}{1-\cos ^{2} x} d x=-\int \frac{d x}{\sin ^{2} x}=\int-\cos e c^{2} x d x=\cot x+c \end{aligned}$
$\begin{aligned} \int \frac{d x}{\cos 2 x-\cos ^{2} x} &=\int \frac{1}{2 \cos ^{2} x-1-\cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x-1} d x \\ &=\int \frac{-1}{1-\cos ^{2} x} d x=-\int \frac{d x}{\sin ^{2} x}=\int-\cos e c^{2} x d x=\cot x+c \end{aligned}$
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