Number of points of discontinuity Find the number of points where 
$$f(\theta)=\int_{-1}^{1}\frac{\sin\theta dx}{1-2x\cos\theta +x^2}$$ is discontinuous where $\theta \in [0,2\pi]$
I am not able to find $f(\theta)$ in terms of $\theta$,$sin\theta$ in the numerator i can take out but $cos\theta$ in the denominator is troublesome.Can someone tell a good way to integate it or some other way to solve it.

Let
$$ f(\theta)=\int_{-1}^{1}\frac{\sin\theta dx}{1-2x\cos\theta +x^2}. $$
If $\theta=0,\pi$ or $2\pi$, there is nothing to do since $f(\theta)=0$. Otherwise, using
$$ \int\frac{1}{(x-a)^2+b^2}dx=\frac{1}{b}\arctan\frac{x-a}{b}, $$
we have
\begin{eqnarray}
f(\theta)&=&\sin\theta\int_{-1}^{1}\frac{dx}{(x-\cos\theta)^2+\sin^2\theta}\\
&=&\arctan\frac{x-\cos\theta}{\sin\theta}\big|_{x=-1}^{x=1}\\
&=&\arctan\frac{1-\cos\theta}{\sin\theta}-\arctan\frac{-1-\cos\theta}{\sin\theta}\\
&=&\arctan(\tan\frac{\theta}{2})+\arctan(\cot\frac{\theta}{2})\\
&=&\frac{\pi}{2}.
\end{eqnarray}
