# If $\alpha$ is an acute angle, show that $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$

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.

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+1: For showing your work. –  Aryabhata Apr 14 '11 at 23:13
If you are familiar with the trig identities (esspecially formula of $\tan(A-B)$) you can also simply calculate $\tan$ of the blue bit, and try to reduce it to $\tan(\frac{1}{2} \alpha)$. Of course, you also need to make sure that the blue bit represents an acute angle.... The solutions listed below are simpler. –  N. S. Apr 15 '11 at 1:45

Use $1 + \cos \alpha = 2\cos^2 (\frac{\alpha}{2})$

and $\sin \alpha = 2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})$

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Got it, thank you! –  Lyrebird Apr 14 '11 at 23:32

Draw the right triangle to see that $\arctan(\cot(\alpha))=\frac{\pi}{2}-\alpha$

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and so also $\arctan(\cot(\alpha /2 ))=\pi-\alpha/2$. Then subtract. –  Henry Apr 14 '11 at 23:20
Many thanks, guys. –  Lyrebird Apr 14 '11 at 23:32
$\arctan\cot\alpha=\frac{1}{2}\pi-\alpha \neq \pi-\alpha$. –  John Bentin Apr 15 '11 at 18:02
@John: corrected. Thanks. –  Ross Millikan Apr 15 '11 at 19:21
Or you can use this: $$\tan^{-1}{x} - \tan^{-1}{y} = \tan^{-1}\biggl(\frac{x-y}{1+xy}\biggr)$$