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प्रश्न

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

\[\text{Let I}= \int\frac{\sin\left( x - a \right)}{\sin\left( x - b \right)}dx\]
\[\text{Putting  x }- b = t \]
\[ \Rightarrow x = b + t\]
\[\text{and}\ dx = dt\]
`∴  I = ∫   sin( b + t - a ) / sin t  dt `
`∴  I = ∫   sin {( b-a )+t } / sin t  dt `


`∴  I = ∫   {sin( b - a )cos t}/sin t  +  ∫   {cos ( b  - a ) sin t} / sin t  dt `


\[ = \int\text{sin}\left( \text{b - a} \right)\text{cot t dt} + \int\text{cos}\left( b - a \right)dt\]
\[ = \text{sin}\left( \text{b - a }\right) \text{ln }\left| \text{sin t} \right| + \text{t }\text{cos}\left( b - a \right) + C\]
\[ = \text{sin}\left( \text{b - a }\right) \text{ln }\left| \text{sin}\left( \text{x - b }\right) \right| + \left( \text{x - b} \right)\text{cos}\left( \text{b - a} \right) + C   \left[ \because t = x - b \right]\]

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अध्याय 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.08 | Q 7 | पृष्ठ ४७

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