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प्रश्न
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
योग
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उत्तर
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right)dx\]
` = ∫ cot ^-1 (( 2 sin x cos x) /( 2 sin^2 x))` dx ` [∴ sin 2x = 2 sin x cos x & 1 - cos 2x = 2 sin^2 x ]`
\[ = \int \cot^{- 1} \left( \cot x \right)dx\]
` = ∫ x dx `
\[ = \frac{x^2}{2} + C\]
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