The value of the integral ${\int }_0^1 \frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{xcos}\mathrm{\alpha }+1}$ is equal to $(\mathrm{\alpha }\in (0,\mathrm{\pi }\left)\right)$

Given integral
$={\int }_0^1 \frac{\mathrm{dx}}{(\mathrm{x}+\cos \mathrm{\alpha }{)}^2+\left(1-{\cos }^2\mathrm{\alpha }\right)}$
$\begin{array}{l}={\int }_0^1 \frac{\mathrm{dx}}{(\mathrm{x}+\cos \mathrm{\alpha }{)}^2+{\sin }^2\mathrm{\alpha }}\\ =\frac{1}{\sin \mathrm{\alpha }}{\left|{\tan }^{-1}\frac{\mathrm{x}+\cos \mathrm{\alpha }}{\sin \mathrm{\alpha }}\right|}_0^1\\ =\frac{1}{\sin \mathrm{\alpha }}\left[{\tan }^{-1}\frac{1+\cos \mathrm{\alpha }}{\sin \mathrm{\alpha }}-{\tan }^{-1}\frac{\cos \mathrm{\alpha }}{\sin \mathrm{\alpha }}\right]\\ =\frac{1}{\sin \mathrm{\alpha }}\left[{\tan }^{-1}\cot \frac{\mathrm{\alpha }}{2}-{\tan }^{-1}(\cot \mathrm{\alpha })\right]\\ =\frac{1}{\sin \mathrm{\alpha }}\left[{\tan }^{-1}\tan \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\alpha }}{2}\right)-{\tan }^{-1}\tan \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)\right]\\ =\frac{1}{\sin \mathrm{\alpha }}\left[\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\alpha }}{2}\right)-\left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)\right]=\frac{\mathrm{\alpha }}{2\sin \mathrm{\alpha }}\end{array}$