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$\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x} d x=$
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The correct answer is:
$\log \left|\tan ^{2} x\right|+c$
$\begin{aligned} I &=\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x}=\int \frac{2 \sin x \cos x d x}{\sin ^{2} x \cos ^{2} x}=2 \int \frac{1}{\sin x \cos x} d x=2 \int \frac{2}{2 \sin x \cos x} d x \\ &=2 \times 2 \int \frac{1}{\sin 2 x} d x=4 \int \operatorname{cosec} 2 x d x \\ &=2 \log |\tan x|+c=\log \left|\tan ^{2} x\right|+c \end{aligned}$
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