# Evaluate: $\int \frac {\sin (x)}{\sin (x-\alpha) } dx$

Evaluate: $$\int \frac {\sin (x)}{\sin (x-\alpha) } dx$$

My Attempt: $$=\int \dfrac {\sin (x)}{\sin (x-\alpha) }dx$$ $$=-\int \dfrac {-\sin (x)}{\sin (x).\cos (\alpha) - \cos (x).\sin (\alpha )} dx$$ Let, $$\cos (x)=t$$ $$-\sin (x).dx=dt$$ Now, $$=-\int \dfrac {dt}{\sqrt {1-t^2}.\cos (\alpha)-t.\sin (\alpha)}$$

Rough reasoning: \begin{align*} \int\dfrac{\sin x}{\sin(x-\alpha)}dx&=\int\dfrac{\sin(u+\alpha)}{\sin u}du\\ &=\int\dfrac{\sin u\cos\alpha+\cos u\sin\alpha}{\sin u}du\\ &=\int\cos\alpha du+\sin\alpha\int\cot udu\\ &=\cdots, \end{align*} where $u=x-\alpha$.
Substitute $z=x-\alpha$ to get $\sin(\alpha +z)$ at the numerator
Then $$I=\int \frac {\sin(z+\alpha)} {\sin(z)}dz$$ $$I=\int \frac {\sin(z)\cos(\alpha)+\sin(\alpha)\cos(z)} {\sin(z)}dz$$ $$I=\sin(\alpha)\int \cot(z)dz+z\cos(\alpha)+K$$ $$\boxed{I=\sin(\alpha)\ln|\sin(x-\alpha)|+x\cos(\alpha)+K}$$
$\int{\frac{\sin(x)}{\sin(x-a)}dx}=\int{\frac{\sin(u+a)}{\sin(u)}du}$ where $u=x-a$. From here you can use the same sin identity from your attempt to obtain $$\int{\frac{\sin(u+a)}{\sin(u)}du}=\int{\frac{\sin(u)\cos(a)+\sin(a)\cos(u)}{\sin(u)}du}=\int{\cos(a)+\sin(a)\cot(u)du}=u\cos(a)+\sin(a)\ln(\sin(u))+C=x\cos(a)+\sin(a)\ln(\sin(x-a))+C$$ where C is some constant.