The value of the integral $\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$, where $0 < \alpha < \dfrac{\pi}{2}$, is equal to

$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$

$\implies \displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +\cos^2 \alpha+\sin^2 \alpha}$

$\implies \displaystyle \int_0^1 \cfrac{dx}{(x+ \cos \alpha)^2 + \sin^2 \alpha}$

$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{x+\cos x}{\sin x}\Bigg]_0^1$ $\left[\because \int \cfrac{1}{x^2+a^2}=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) \right]$

$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{1+\cos \alpha}{\sin \alpha} - \tan^{-1} \cfrac{\cos \alpha}{\sin \alpha}\Bigg]$

$\implies \cfrac{1}{\sin \alpha} \Bigg[ \tan^{-1} \cfrac{\cfrac{1+\cos \alpha}{\sin \alpha} - \cfrac{\cos \alpha}{\sin \alpha}}{1+\cfrac{\cos \alpha + \cos^2 \alpha}{\sin^2 \alpha}}\Bigg]$

$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \cfrac{\sin \alpha}{1+\cos \alpha}$

$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \tan \cfrac{\alpha}{2}$

$\implies \cfrac{\alpha}{2 \sin \alpha}$