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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.

  • 0
    I have changed the formatting of the title so as to [make it take up less vertical space](https://math.meta.stackexchange.com/a/9686/290189) -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See [here](https://math.meta.stackexchange.com/a/9730) for more information. Please take this into consideration for future questions. Thanks in advance.2018-03-12

3 Answers 3

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

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

12

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

  • 1
    $\arctan\cot\alpha=\frac{1}{2}\pi-\alpha \neq \pi-\alpha$.2011-04-15
9

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.