Evaluating $\int\frac{\sin x}{\sin (x-a)\cdot \sin (x-b)}\,\mathrm dx$

Question

$$\int\frac{\sin x}{\sin (x-a)\cdot \sin (x-b)}\,\mathrm dx$$

My Try::

\begin{align} &\displaystyle \frac{1}{\sin (b-a)}\int\frac{\sin \{(x-a)-(x-b)\}}{\sin (x-a)\cdot \sin (x-b)}\cdot\sin x\,\mathrm dx\\ =& \frac{1}{\sin (b-a)}\int \left\{\cot (x-b)-\cot (x-a)\right\}\cdot \sin x \,\mathrm dx \end{align}

Now how can I proceed after that?

Answer 1

Wrong answer for now.

First let's rewrite: $$\frac{1}{\sin(x-a)\sin(x-b)}=\frac{p}{\sin(x-a)}+\frac{q}{\sin(x-b)} \quad(1)$$ Now multiplying $(1)$ by $\sin(x-a)$ followed by letting $x=a$ yields: $$\left(\frac{1}{\sin(x-b)}=p+q\underset{\rightarrow 0}{\frac{\sin(x-a)}{\sin(x-b)}}\right)\bigg|_{x=a}\Rightarrow p=\frac{1}{\sin(a-b)}$$ Similarly to obtain q, multiply $(1)$ by $\sin(x-b)$ and plugg $x=b$ gives: $$\left(\frac{1}{\sin(x-a)}=p\underset{\rightarrow 0}{\frac{\sin(x-b)}{\sin(x-a)}}+q\right)\bigg|_{x=b}\Rightarrow q=\frac{1}{\sin(b-a)}=-p$$ Back to the original integral, we have: $$I=\int\frac{\sin x}{\sin (x-a)\cdot \sin (x-b)}dx=\frac{1}{\sin(a-b)}\left(\int\frac{\sin x}{\sin (x-a)}dx-\int\frac{\sin x}{\sin(x-b)}dx\right)$$ Now, it's enough to solve only one of these integrals. $$\int\frac{\sin x}{\sin(x-t)}dx=\int\frac{\sin(x-t+t)}{\sin (x-t)}dx=\cos t\int\frac{\sin(x-t)}{\sin(x-t)}dx+\sin t\int\frac{\cos(x-t)}{\sin(x-t)}dx=$$ $$=\cos t \cdot x +\sin t \cdot \ln|\sin(x-t)|+c$$ Thus replacing $t$ by $a$ and $b$ we get the answer to be: $$I=\frac{1}{\sin(a-b)}\left((\cos a-\cos b)x+\sin a \ln|\sin(x-a)|-\sin b \ln|\sin(x-b)|\right)+C$$