Question
Evaluate:
$\int \frac{\cos 2 x}{\sin ^2 x} \cdot d x$

Answer

$ \int \frac{\cos 2 x}{\sin ^2 x} d x=\int \frac{\left(1-2 \sin ^2 x\right)}{\sin ^2 x} d x$
$=\int\left(\frac{1}{\sin ^2 x}-\frac{2 \sin ^2 x}{\sin ^2 x}\right) d x$
$=\int \operatorname{cosec}^2 x d x-2 \int d x$
$=-\cot x-2 x+c .$

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