Question:
Consider
$$F(x) = \frac{1}{\sin(x-a) \ \sin(x-b) \ \sin(x-c)}$$
Then, how to compute $\int F(x) \, \mathrm{d}x$?
Edit:
I have tried what I know about integral solving methods. I have tried doing it by parts taking second function as 1 it turned out to be a mess. I tried multiplying and dividing by constants having sin function in tricky ways but still a mess.
Following @acchilehui's suggestion, take advantage of the fact:
$$\sin{(x-\theta)} = \sin{x}\cos{\theta} - \cos{x} \sin{\theta},$$
and make use of the Weierstrass substitution to write your integral as follows:
It seems that the denominator can be nicely factorized up as follows (perhabps this is the tricky part):
$$\text{Den} = (t+\cot{a/2})(t+\cot{b/2})(t+\cot{c/2})(t-\tan{a/2})(t-\tan{b/2})(t-\tan{c/2})$$
so you can perform a partial fraction decomposition as follows:
$$ I = \int Q(t) \, \mathrm{d}t = \int \sum_i \frac{A_i}{t-t_i} \, \mathrm{d}t,$$
where $t_i$ are the roots of $\text{Den}$ and $A_i$ are given by (see here to see how it works):
$$A_i = \lim_{t\to t_i} Q(t)(t-t_i), $$
which leads to the solution
Substituting back $t = \tan{x/2}$ gives you the result in terms of $x$. Hope this helps!
Cheers.