Integral of $\int_0^1\frac{1}{x^2+2x\cos\alpha+1}dx$.

Question

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 : $$ I=\frac{\alpha}{2 \sin(\alpha)}$$

Answer 1

Hint

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

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

Answer 2

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.

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Another Approach

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