Explanation

(d)
Let $I = \int_{}^{}\frac{\sin x}{\sin{(x - \alpha)}}dx$
Put $x - \alpha = t\ \ \ \Rightarrow \ \ \ dx = dt$
$\therefore\ \ \ \ I = \frac{\sin{(\alpha + t)}}{\sin t}dt$
$= \int_{}^{}\frac{\sin\alpha\cot t + \cos\alpha\sin t}{\sin t}dt$
$= \sin{\alpha\int_{}^{}{\cot tdt + \cos{\alpha\int_{}^{}{dt}}}}$
$= \sin\alpha\log{\sin t + + \cos\alpha.t + c_{1}}$
$= \sin{\alpha\log{\sin{(x - \alpha)}}} + \cos\alpha.(x - \alpha) + c_{1}$
$= \sin\alpha\log{\sin(x - \alpha)} + x\cos\alpha + c$
[let $c = - \alpha\cos{\alpha + c_{1}}$]