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