The value of $\int \frac{\sin x}{\sin (x-\pi /4)}dx$ is:

Let $x-\pi /4=t⟹x=t+\pi /4$. Then $dx=dt$.The integral becomes:
$$\int \frac{\sin (t+\pi /4)}{\sin t}dt=\int \frac{\sin t\cos (\pi /4)+\cos t\sin (\pi /4)}{\sin t}dt$$
$$=\int \left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cot t\right]dt$$
$$=\frac{1}{\sqrt{2}}t+\frac{1}{\sqrt{2}}\ln |\sin t|+C^{\prime }$$
Substituting $t=x-\pi /4$:
$$=\frac{1}{\sqrt{2}}(x-\pi /4)+\frac{1}{\sqrt{2}}\ln |\sin (x-\pi /4)|+C^{\prime }$$
However, looking at the provided answer format, the factor $1/\sqrt{2}$ is handled differently. Let's re-examine the expression:$\frac{\sin x}{\sin (x-\pi /4)}=\frac{\sin ((x-\pi /4)+\pi /4)}{\sin (x-\pi /4)}=\cos (\pi /4)+\sin (\pi /4)\cot (x-\pi /4)$.The integral is $\frac{1}{\sqrt{2}}[x+\ln \sin (x-\pi /4)]+C$.