Evaluate the following integrals:


Evaluate: $\int \frac{2}{1-\cos 2 x} d x$


Given, $\int \frac{2}{1-\cos 2 x} d x$

We Know that, $\cos 2 x=1-2 \sin ^{2} x$

$\Rightarrow 1-\cos 2 x=2 \sin ^{2} x$

Substitute this in the given,

$=\int \frac{2}{2 \sin ^{2} x} d x$

$=\int \frac{1}{\sin ^{2} x} d x$

$=\int \operatorname{cosec}^{2} x d x$

$=-\cot x+c$

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