Integral of $\int_0^1\frac{1}{x^2+2x\cos\alpha+1}dx$. 
Prove that :
     $$I=\int_0^1\frac{1}{x^2+2x\cos\alpha+1}dx=\frac{\alpha}{2\sin(\alpha)}$$

I can find this result:
$$ I=\frac{1}{\sin(\alpha)}\left(\arctan\left(\frac{1+\cos(\alpha)}{\sin(\alpha)}\right)-\arctan\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)\right) $$ 
but I don't now how to prove that :
$$ I=\frac{\alpha}{2 \sin(\alpha)}$$
 A: Hint
Using
$$x^2+2x\cos\alpha +1=(x-\cos\alpha )^2+\sin^2\alpha  ,$$
you get
$$I=\frac{1}{\sin^2 \alpha }\int_0^1 \frac{1}{\left(\frac{x-\cos \alpha }{\sin \alpha }\right)^2+1}dx.$$
A: Possible Continuation
$$
\begin{align}
\frac1{\sin(\alpha)}\left(\arctan\left(\frac{1+\cos(\alpha)}{\sin(\alpha)}\right)-\arctan\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)\right)
&=\frac1{\sin(\alpha)}\arctan\left(\frac{\frac{1+\cos(\alpha)}{\sin(\alpha)}-\frac{\cos(\alpha)}{\sin(\alpha)}}{1+\frac{1+\cos(\alpha)}{\sin(\alpha)}\frac{\cos(\alpha)}{\sin(\alpha)}}\right)\\
&=\frac1{\sin(\alpha)}\arctan\left(\frac{\sin(\alpha)}{1+\cos(\alpha)}\right)\\[6pt]
&=\frac{\alpha}{2\sin(\alpha)}
\end{align}
$$
where we use $\tan\left(\frac\alpha2\right)=\frac{\sin(\alpha)}{1+\cos(\alpha)}$ as shown in this answer.

Another Approach
Using partial fractions, we have
$$
\begin{align}
\int_0^1\frac1{x^2+2x\cos(\alpha)+1}\,\mathrm{d}x
&=\frac1{2i\sin(\alpha)}\int_0^1\left(\frac1{x+e^{-i\alpha}}-\frac1{x+e^{i\alpha}}\right)\,\mathrm{d}x\\
&=\frac1{2i\sin(\alpha)}\left[\log\left(\frac{1+e^{-i\alpha}}{1+e^{i\alpha}}\right)-\log\left(\frac{e^{-i\alpha}}{e^{i\alpha}}\right)\right]\\[3pt]
&=\frac{\alpha}{2\sin(\alpha)}
\end{align}
$$
