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Prove that : $$I=\int_0^1\frac{1}{x^2+2x\cos\alpha+1}dx=\frac{\alpha}{2\sin(\alpha)}$$

I can find this result:

$$ I=\frac{1}{\sin(\alpha)}\left(\arctan\left(\frac{1+\cos(\alpha)}{\sin(\alpha)}\right)-\arctan\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)\right) $$

but I don't now how to prove that : $$ I=\frac{\alpha}{2 \sin(\alpha)}$$

Delta-u
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user574907
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  • I would guess to trig +1 into $\sin(\alpha)^2 + \cos(\alpha)^2$ and use it to complete the square. – mathreadler Jul 08 '18 at 18:08
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    Buy yourself a copy of Nahin's Inside Interesting Integrals ... by the time you have worked through chapter 3 you will know exactly how to do this particular integral – Mandelbrot Jul 08 '18 at 18:10
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    $$\tan\left(\arctan(\frac{1+\cos\alpha}{\sin\alpha})-\arctan(\frac{\cos\alpha}{\sin\alpha})\right) = \frac{sin(\alpha)}{1+cos(\alpha)}= \frac{2sin(\alpha/2)\cos(\alpha/2)}{2cos^2(\alpha/2)}=\tan(\alpha/2)$$ then $$\arctan(\frac{1+cos(\alpha)}{sin(\alpha)})-Arctan(\frac{cos(\alpha)}{sin(\alpha)})=\dfrac{\alpha}{2}$$ – Nosrati Jul 08 '18 at 18:26

2 Answers2

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Hint

Using $$x^2+2x\cos\alpha +1=(x-\cos\alpha )^2+\sin^2\alpha ,$$

you get $$I=\frac{1}{\sin^2 \alpha }\int_0^1 \frac{1}{\left(\frac{x-\cos \alpha }{\sin \alpha }\right)^2+1}dx.$$

Surb
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Possible Continuation $$ \begin{align} \frac1{\sin(\alpha)}\left(\arctan\left(\frac{1+\cos(\alpha)}{\sin(\alpha)}\right)-\arctan\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)\right) &=\frac1{\sin(\alpha)}\arctan\left(\frac{\frac{1+\cos(\alpha)}{\sin(\alpha)}-\frac{\cos(\alpha)}{\sin(\alpha)}}{1+\frac{1+\cos(\alpha)}{\sin(\alpha)}\frac{\cos(\alpha)}{\sin(\alpha)}}\right)\\ &=\frac1{\sin(\alpha)}\arctan\left(\frac{\sin(\alpha)}{1+\cos(\alpha)}\right)\\[6pt] &=\frac{\alpha}{2\sin(\alpha)} \end{align} $$ where we use $\tan\left(\frac\alpha2\right)=\frac{\sin(\alpha)}{1+\cos(\alpha)}$ as shown in this answer.

Another Approach

Using partial fractions, we have $$ \begin{align} \int_0^1\frac1{x^2+2x\cos(\alpha)+1}\,\mathrm{d}x &=\frac1{2i\sin(\alpha)}\int_0^1\left(\frac1{x+e^{-i\alpha}}-\frac1{x+e^{i\alpha}}\right)\,\mathrm{d}x\\ &=\frac1{2i\sin(\alpha)}\left[\log\left(\frac{1+e^{-i\alpha}}{1+e^{i\alpha}}\right)-\log\left(\frac{e^{-i\alpha}}{e^{i\alpha}}\right)\right]\\[3pt] &=\frac{\alpha}{2\sin(\alpha)} \end{align} $$

robjohn
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  • This comment is misplaced - I can't put it where it belongs. Why did you delete your answer to that other question? Your answer and mine are the two good ones! (The answer to the comment asking why we can't do the same with that other integral is we can, why do you think we can't?) – David C. Ullrich Jul 15 '18 at 17:42
  • @DavidC.Ullrich: I had deleted it because the question had very little context. I undeleted it 22 seconds later because I saw several other answers and I figured that mine was a better alternative to most (but not all ;-). In light of multiple existing answers, the punitive effect of a non-answer (to discourage PSQs) seemed to be outweighed by the benefit of an answer. – robjohn Jul 15 '18 at 18:25