Angular integrals used in QED I am reading a research paper
 and am stuck at a point where the author uses angular integrals.  I don't have any idea about it and would like help. 
 The angular integral is:
$$I_k (y)=\int_0^\pi  {\sin^2(\theta)\cos^{2k}(\theta) \over \beta^2(y)-\cos^2 (\theta)} d\theta$$
The author directly give results to this integral without proof.
The results are:
$$I_0 (y)=\pi \left(1-\sqrt{1-{1\over\beta^2 (y)} }  \right),$$
$$I_1 (y)={-\pi \over 2}+\beta^2 (y)I_0 (y),$$
I don't know how he got these results.  Can anyone please help me?

An interesting approach to these integrals is the use of complex integration. In this case, we can use the fact that
with
\begin{equation}
f(\sin\theta,\cos\theta) = \frac{\sin^2\theta \cos^{2k} \theta}{\beta^2-\cos^2\theta}
\end{equation}
we have
\begin{eqnarray}
\int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) &=&
\frac{1}{2} \int_0^{2\pi} \mathrm{d}\theta \,f(\sin\theta,\cos\theta)
\\
&=& \frac{1}{2} \int_{|z|=1} \frac{\mathrm{d} z}{i z} \, f\left(\frac{z-z^{-1}}{2i},\frac{z+z^{-1}}{2}\right)
\\
&=&  - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} (z^4-4 \beta^2 z^2+2z^2+1)} 
\\
&=&  - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)}  \, ,
\end{eqnarray}
with $z_i$ the simple poles
\begin{eqnarray}
z_1 &=& -\beta - \sqrt{\beta^2 - 1}
\\ 
z_2 &=& \beta - \sqrt{\beta^2 - 1}
\\ 
z_3 &=& -\beta + \sqrt{\beta^2 - 1}
\\ 
z_4 &=& \beta + \sqrt{\beta^2 - 1} \,.
\end{eqnarray}
By looking at the result, seems like the author assumes the range of $\beta$ is such that only $z_2$ and $z_3$ lay inside the unit circle (besides the $(2k+1)$-order pole at the origin). Add the contributions from those 3 poles and you can actually find a formula for arbitrary $k$ with relative ease.
More precisely, if we write
\begin{equation}
g = \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)} \,
\end{equation}
then we find
\begin{eqnarray}
\int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) = && \frac{\pi}{2^{2k}} \left\{ \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[z^{2k+1} g\right] + \lim_{z\rightarrow z_2} \left[(z-z_2)g\right] + \lim_{z\rightarrow z_3} \left[(z-z_3)g\right] \right\} \, .
\end{eqnarray}
Replacing and simplifying gives us
\begin{equation}
I_k(y) = \frac{\pi}{2^{2k}} \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[ \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^4+z^2\left[2-4\beta^2(y)\right] + 1} \right] - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, .
\end{equation}
Finally, when expanding the first term, it turns out this can be further simplified, producing
\begin{equation}
I_k(y) = - \pi \sum_{m=0}^k \frac{(2m-3)!!}{(2m)!!} \beta^{2k-2m}(y) - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, ,
\end{equation}
where we take $(-3)!! = -1$, $(-1)!! = 1$ and $0!! = 1$. This also allows us to generalize the relation between successive values of $I_k(y)$, where we have
\begin{equation}
I_k(y) = - \frac{(2k-3)!!}{(2k)!!} \pi + \beta^{2}(y) I_{k-1}(y) \,.
\end{equation}
In hindsight, this result could have been deduced in a couple of lines, by noticing that the difference between the integrands of $I_k(y)$ and $\beta^2(y) I_{k-1}(y)$ simplifies to
\begin{equation}
\frac{\sin^2\theta \cos^{2k} \theta}{\beta^2(y)-\cos^2\theta} - \beta^2(y) \frac{\sin^2\theta \cos^{2k-2} \theta}{\beta^2(y)-\cos^2\theta} = \cos^{2k} \theta - \cos^{2k-2} \theta \,.
\end{equation}
Hence
\begin{eqnarray}
I_k(y) - \beta^2(y) I_{k-1}(y) &=& \int_0^\pi \mathrm{d}\theta \left[ \cos^{2k} \theta - \cos^{2k-2} \theta \right]
\\
&=& \pi \left[ \frac{(2k-1)!!}{(2k)!!} - \frac{(2k-3)!!}{(2k-2)!!} \right]
\\
&=& - \frac{(2k-3)!!}{(2k)!!} \pi \, ,
\end{eqnarray}
for $k \geq 1$.
