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