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प्रश्न
\[\int\frac{1}{1 - \cos 2x} dx\]
बेरीज
उत्तर
\[\int\frac{dx}{1 - \cos \left( 2x \right)} \left[ \therefore 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \right]\]
\[ = \int\frac{dx}{2 \sin^2 x}\]
\[ = \frac{1}{2}\int {cosec}^2 x dx\]
\[ = \frac{1}{2}\left[ - \cot x \right] + C\]
\[ = - \frac{1}{2}\cot x + C\]
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