TITLE: If $\alpha$ is an acute angle, show that $\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
QUESTION [26 upvotes]: If $\alpha$ is an acute angle, show that $\displaystyle  \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
My attempt: 
Write $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\alpha} = (x+\cos{\alpha})^2+\sin^2{\alpha}$, we have:
$\displaystyle  \begin{aligned}\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} & = \int_0^1 \frac{dx}{(x+\cos{\alpha})^2+\sin^2{\alpha}} = \bigg[\frac{1}{\sin{\alpha}}\tan^{-1}\left(\frac{x+\cos{\alpha}}{\sin{\alpha}}\right)\bigg]_{x=0}^1 \\ & = \frac{1}{\sin{\alpha}}\bigg[\color{blue}{\tan^{-1}\left(\frac{1+\cos{\alpha}}{\sin{\alpha}}\right)-\tan^{-1}\left(\frac{\cos{\alpha}}{\sin{\alpha}}\right)}\bigg] \end{aligned}$
I'm not sure, however, how the blue bit reduces to $\frac{1}{2}\alpha$. Any hints/suggestions? Thanks.

REPLY [9 votes]: Or you can use this: $$\tan^{-1}{x} - \tan^{-1}{y} = \tan^{-1}\biggl(\frac{x-y}{1+xy}\biggr)$$
I am doing just the calculation part. We have 
\begin{align*} \frac{1}{1+ \frac{\cos\alpha+\cos^{2}\alpha}{\sin^{2}\alpha}}\times \frac{1 + \cos\alpha}{\sin\alpha} - \frac{\cos\alpha}{\sin\alpha} &= \frac{1}{\sin\alpha} \times \frac{\sin^{2}\alpha}{\sin^{2}\alpha + \cos^{2}\alpha + \cos\alpha} \\ &= \frac{\sin\alpha}{1 + \cos \alpha} \\ &=\frac{ \displaystyle 2 \sin\frac{\alpha}{2}\cdot \cos\frac{\alpha}{2}}{2 \cos^{2}\frac{\alpha}{2}} \\ &=\tan\frac{\alpha}{2}
\end{align*}
Now take the inverse and then get the answer.