Question

Integration of sin x minus alpha upon sin x + alpha

Answer 1

Answer:

$\int \frac{ \sin(x - \alpha ) }{ \sin(x + \alpha ) } dx \\ \: \\ put \: \: y = x + \alpha \: \implies \: dy = dx \\ \\ \int \frac{ \sin(y -2 \alpha ) }{ \sin( y) } dy \\ \\ = \int \: \frac{ \sin(y) \cos(2 \alpha ) - \cos(y) \sin(2 \alpha ) }{ \sin(y) } dy \\ \\ = \int \: ( \cos(2 \alpha ) - \sin(2 \alpha ) \cot(y) )dy \\ \\ = y \cos(2 \alpha ) - \sin(2 \alpha ) ln( \sin(y) ) + c \\ \\ = (x + \alpha ) \cos(2 \alpha ) - \sin(2 \alpha ) ln( \sin(x + \alpha ) ) + C$