Question

\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]

Answer 1

\[\text{Let I }= \int\frac{\sin\left( x - \alpha \right)}{\sin\left( x + \alpha \right)}dx\]

\[\text{Putting x} + \alpha = t \]

\[ \Rightarrow x = t - \alpha\]

\[\text{and}\ dx = dt\]

\[ \therefore I = \int\frac{\sin \left( t - 2\alpha \right)}{\sin t}dt\]

\[ = \int\left( \frac{\sin t \cos 2\alpha}{\sin t} - \frac{\cos t \sin 2\alpha}{\sin t} \right)dt\]

\[ = \cos 2\alpha\  ∫ dt - \sin 2\alpha\int\text{cot t dt}\]

\[ = t\cos 2\alpha - \text{sin 2}\alpha \text{ln }\left| \sin t \right| + C\]

\[ = \left( x + \alpha \right)\text{cos 2}\alpha - \text{sin 2}\alpha \text{ln }\left| \sin \left( x + \alpha \right) \right| + C \left[ \because t = x + \alpha \right]\]

\[ = x\cos 2\alpha - \text{sin 2}\alpha \text{ln }\left| \sin \left( x + \alpha \right) \right| + C\]