How to calculate the following integral if $\varepsilon \in (0,1)$: $$\int \limits_{0}^{\pi}\frac{d\varphi}{(1+\varepsilon\cos \varphi)^2}$$
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1Similar: https://math.stackexchange.com/questions/1846774/how-to-evaluate-int-02-pi-frac11a-cos-theta2-d-theta-without-c – StubbornAtom Sep 28 '17 at 18:11
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this says WA$$\text{ConditionalExpression}\left[\frac{\log \left(\frac{1-\epsilon }{\sqrt{\epsilon ^2-1}}\right)-\log \left(\frac{\epsilon -1}{\sqrt{\epsilon ^2-1}}\right)}{\left(\epsilon ^2-1\right)^{3/2}},\left(\Re\left(\sec ^{-1}(\epsilon )\right)<0\lor \Re\left(\sec ^{-1}(-\epsilon )\right)<0\lor \csc ^{-1}(\epsilon )\notin \mathbb{R}\right)\land ((\Re(\epsilon )+1>0\land \Re(\epsilon )<1\land \Re(\epsilon )\neq 0)\lor \epsilon \notin \mathbb{R})\right]$$ – Dr. Sonnhard Graubner Sep 28 '17 at 18:13
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Possible duplicate of How to evaluate $\int_0^{2\pi}\frac{1}{(1+a\cos {\theta})^2},d\theta$ without contour integration? – Mårten W Sep 28 '17 at 18:18
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Can you recognize the area enclosed by an ellipse? – Jack D'Aurizio Sep 28 '17 at 19:50
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Hint: Use the substitution $$ s=\tan{\frac{\varphi}{2}}, \quad \sin{\varphi}=\frac{2s}{s^2+1}, \quad \cos{\varphi} = \frac{1-s^2}{s^2+1}. $$
(Apparently more details are given in this answer.)
Mårten W
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I have applied trigonometric substitution and I have got the following integral $\int \limits_{0}^{\infty}\dfrac{(1+u^2)du}{(1+u^2+\varepsilon-\varepsilon u^2)^2}$ and I stuck – RFZ Sep 29 '17 at 10:36
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@RFZ: You're missing a factor two in the numerator, but otherwise it looks good. Next thing to do is bring the denominator to the form $(1+v^2)^2$. Do you see what substitution accomplishes this (there are two)? – Mårten W Sep 29 '17 at 11:28
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Complex analysis approach: by letting $z=e^{i\varphi}$ we obtain \begin{align*} \int_{0}^{\pi}\frac{d\varphi}{(1+\varepsilon\cos \varphi)^2}&= \frac{1}{2}\int \limits_{-\pi}^{\pi}\frac{d\varphi}{(1+\varepsilon\cos \varphi)^2}\\ &=\frac{1}{2}\int \limits_{|z|=1}\frac{1}{(1+\varepsilon(z+1/z)/2)^2}\cdot \frac{dz}{iz}\\ &=\frac{2}{i}\int \limits_{|z|=1}\frac{z}{(\varepsilon z^2+2z+\varepsilon)^2}\,dz\\ &=\frac{4\pi}{\varepsilon^2}\,\mbox{Res}\left(\frac{z}{ (z-w_+)^2(z-w_-)^2},w_+ \right)\\ &=\frac{\pi}{(1-\varepsilon^2)^{3/2}}. \end{align*} where $\displaystyle w_{\pm}=\frac{\pm\sqrt{1-\varepsilon^2}-1}{\varepsilon}$.
Robert Z
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$$\int \frac{d\varphi}{(1+\varepsilon\cos\varphi)^2} = -\frac{1}{(1-\varepsilon^2)}\Bigg[\frac{\varepsilon\sin\varphi}{1+\varepsilon\cos\varphi}-\int \frac{d\varphi}{1+\varepsilon\cos\varphi}\Bigg]$$
Table of Integrals, Series, And Products 7th ed 2.554.3
G_D
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Your integral does not correspond to the one in the question. The entire denominator should be squared, not just the cosine. – Mårten W Sep 28 '17 at 18:17
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