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∫ 1 Sin ( X − a ) Sin ( X − B ) D X - Mathematics

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प्रश्न

\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
बेरीज
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उत्तर

\[\int\frac{1}{\text{ sin} \left( x - a \right) \cdot \text{ sin}\left( x - b \right)}dx\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\frac{\text{ sin}\left( b - a \right)}{\text{ sin}\left( x - a \right) \cdot \text{ sin }\left( x - b \right)} \text{ dx }\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\frac{\text{ sin}\left[ \left( x - a \right) - \left( x - b \right) \right]}{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} \text{ dx }\]
\[ = \frac{1}{\text{ sin }\left( b - a \right)}\int\frac{\text{ sin }\left( x - a \right) \cdot \cos \left( x - b \right) - \text{ cos} \left( x - a \right) \text{ sin}\left( x - b \right)}{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} \text{ dx }\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\left[ \frac{\text{ sin}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)}{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} - \frac{\text{ cos}\left( x - a \right) \text{ sin}\left( x - b \right)}{\sin \left( x - a \right) \text{ sin}\left( x - b \right)} \right] \text{ dx }\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\left[ \text{ cot}\left( x - b \right) - \text{ cot}\left( x - a \right) \right] \text{ dx }\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\text{ cot}\left( x - b \right) dx - \int\text{ cot}\left( x - a \right) \text{ dx }\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\left[ \text{ ln }\left| \text{ sin}\left( x - b \right) \right| - \text{ ln} \left| \text{ sin}\left( x - a \right) \right| \right] + C\]
\[ = \frac{1}{\text{ sin}\left( b - a \right)}\left[ \text{ ln }\left| \frac{\text{ sin}\left( x - b \right)}{\text{ sin}\left( x - a \right)} \right| \right] + C\]
\[ = \frac{- 1}{\text{ sin}\left( a - b \right)}\left[ \text{ ln}\left| \frac{\text{ sin}\left( x - b \right)}{\text{ sin}\left( x - a \right)} \right| \right] + C\]
\[ = \frac{1}{\text{ sin}\left( a - b \right)} \text{ ln }\left| \frac{\text{ sin}\left( x - a \right)}{\sin \left( x - b \right)} \right| + C\]

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पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 24 | पृष्ठ २०३

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