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प्रश्न
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उत्तर
\[\text{Let I }= \int\frac{\sin\left( x - \alpha \right)}{\sin\left( x + \alpha \right)}dx\]
\[\text{Putting x} + \alpha = t \]
\[ \Rightarrow x = t - \alpha\]
\[\text{and}\ dx = dt\]
\[ \therefore I = \int\frac{\sin \left( t - 2\alpha \right)}{\sin t}dt\]
\[ = \int\left( \frac{\sin t \cos 2\alpha}{\sin t} - \frac{\cos t \sin 2\alpha}{\sin t} \right)dt\]
\[ = \cos 2\alpha\ ∫ dt - \sin 2\alpha\int\text{cot t dt}\]
\[ = t\cos 2\alpha - \text{sin 2}\alpha \text{ln }\left| \sin t \right| + C\]
\[ = \left( x + \alpha \right)\text{cos 2}\alpha - \text{sin 2}\alpha \text{ln }\left| \sin \left( x + \alpha \right) \right| + C \left[ \because t = x + \alpha \right]\]
\[ = x\cos 2\alpha - \text{sin 2}\alpha \text{ln }\left| \sin \left( x + \alpha \right) \right| + C\]
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