Evaluation of $I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$ Evaluate the given integral:
$$I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$$
When it was $I=\int \frac{1}{\sin (x-a) \sin(x-b)}.dx$, I solved it by multiplying and dividing by $\sin (b-a)$. But I am not getting how to proceed here..
Could someone help me with this?

I'd like to expand on @Kajelad explanation. The system can be solved as follows.
We want to perform the decomposition
$$\frac{1}{\sin(x-a)\sin(x-b)\sin(x-c)} = \frac{A}{\sin(x-a)} + \frac{B}{\sin(x-b)} + \frac{C}{\sin(x-c)}$$
or by rewriting we obtain
$$1 = A\sin(x-b)\sin(x-c) + B\sin(x-a)\sin(x-c) + C\sin(x-a)\sin(x-b).$$
As we want this to hold for all $x$ we may select a few values to see if we get a solution. Naturally, the equation should hold if $x=a$ hence we obtain
$$1 = A\sin(a-b)\sin(a-c) + B\sin(0)\sin(a-c) + C\sin(0)\sin(a-b)$$
but as $\sin(0) = 0$ then 
$$A = \frac{1}{\sin(a-b)\sin(a-c)}.$$
Similarly, by selecting $x=b$ and $x=c$ we obtain
$$B = \frac{1}{\sin(b-a)\sin(b-c)}$$
$$C = \frac{1}{\sin(c-a)\sin(c-b)}.$$
Through brute work and using the odd property of $\sin$ and trigonometric formulas these should verify that the decomposition is justified. If so, then the solution is given by
$$\int \left(A\csc(x-a) + B\csc(x-b) + C\csc(x-c)\right) dx$$
$$ = \frac{\ln(\tan(x-a))}{\sin(a-b)\sin(a-c)} + \frac{\ln(\tan(x-b))}{\sin(b-a)\sin(b-c)} + \frac{\ln(\tan(x-c))}{\sin(c-a)\sin(c-b)} + C$$
