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प्रश्न

\[\int\frac{\sin^2 x}{1 + \cos x}   \text{dx} \]
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उत्तर

\[\int\frac{\sin^2 x}{1 + \cos x}dx\]
\[ = \int\frac{\left( 1 - \cos^2 x \right)}{\left( 1 + \cos x \right)}dx\]
\[ = \int\frac{\left( 1 - \cos x \right) \left( 1 + \cos x \right)}{\left( 1 + \cos x \right)}dx\]
\[ = \int \left( 1 - \cos x \right)dx\]
\[ = x - \sin x + C\]

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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 21 | पृष्ठ १५

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