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प्रश्न

\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
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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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अध्याय 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.02 | Q 37 | पृष्ठ १५

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