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प्रश्न
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
बेरीज
उत्तर
\[\int \frac{dx}{1 - \sin\left( \frac{x}{2} \right)}\]
\[ = \int\frac{\left( 1 + \sin \frac{x}{2} \right)}{\left( 1 - \sin \frac{x}{2} \right) \left( 1 + \sin \frac{x}{2} \right)} dx\]
\[ = \int\left( \frac{1 + \sin \frac{x}{2}}{1 - \sin^2 \frac{x}{2}} \right)dx\]
\[ = \int\left( \frac{1 + \sin\frac{x}{2}}{\cos^2 \frac{x}{2}} \right) dx\]
\[ = \int\left( \sec^2 \frac{x}{2} + \sec \frac{x}{2} \text{tan }\frac{x}{2} \right)dx\]
\[ = \frac{\tan \left( \frac{x}{2} \right)}{\frac{1}{2}} + \frac{\sec \left( \frac{x}{2} \right)}{\frac{1}{2}} + C\]
\[ = 2 \left( \tan \frac{x}{2} + \sec \frac{x}{2} \right) + C\]
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