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∫ 1 1 − Sin X 2 D X - Mathematics

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प्रश्न

\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
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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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अध्याय 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.03 | Q 12 | पृष्ठ २३

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