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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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    $\begingroup$ 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. $\endgroup$
    – N. S.
    Apr 15 '11 at 1:45
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ Mar 12 '18 at 16:56
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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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Draw the right triangle to see that $\arctan(\cot(\alpha))=\frac{\pi}{2}-\alpha$

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    $\begingroup$ and so also $\arctan(\cot(\alpha /2 ))=\pi-\alpha/2$. Then subtract. $\endgroup$
    – Henry
    Apr 14 '11 at 23:20
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    $\begingroup$ $\arctan\cot\alpha=\frac{1}{2}\pi-\alpha \neq \pi-\alpha$. $\endgroup$ Apr 15 '11 at 18:02
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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.

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