If $a>b>0$, prove that :

$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$

  • 1
    $\begingroup$ What does the expression on the right mean? $\endgroup$ – L. F. Mar 10 '13 at 19:00
  • $\begingroup$ And why is there a $c$? $\endgroup$ – MITjanitor Mar 10 '13 at 19:01
  • 1
    $\begingroup$ why is it called complex analysis $\endgroup$ – Dominic Michaelis Mar 10 '13 at 19:02
  • $\begingroup$ sorry...edited.. $\endgroup$ – ochapuspita Mar 10 '13 at 19:03
  • $\begingroup$ it use residue theorm... $\endgroup$ – ochapuspita Mar 10 '13 at 19:04

$$I = \int_0^{2 \pi} \dfrac{\sin^2(x)}{a+b \cos(x)} dx \implies aI = \int_0^{2 \pi} \dfrac{\sin^2(x)}{1+\dfrac{b \cos(x)}a}dx$$ \begin{align} aI & = \sum_{k=0}^{\infty}\left(\dfrac{(-b)^k}{a^k} \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx \right) = \sum_{k=0}^{\infty}\left(\dfrac{b^{2k}}{a^{2k}} \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx \right) \end{align} Note that we have thrown away the odd terms since for $k$ odd, the integral $\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx$ is zero. \begin{align} \dfrac{\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx}4 & = \int_0^{\pi/2}\sin^2(x) \cos^{2k}(x) dx\\ & \int_0^{\pi/2}\cos^{2k}(x) dx - \int_0^{\pi/2}\cos^{2k+2}(x) dx\\ & = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2 - \dfrac{2k+1}{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\ & = \dfrac1{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\ & = \dfrac{\pi}{2^{k+2}} \dfrac{(2k-1)(2k-3)\cdots3 \cdot 1}{(k+1)!} = \dfrac{\pi}{2^{2k+2}} \dfrac{(2k)!}{k! (k+1)!} \end{align} Hence, $$\dfrac{aI}{\pi} = \sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}}$$ Now $$\sum_{k=0}^{\infty} \dfrac{\dbinom{2k}k x^{k}}{k+1} = \dfrac{1-\sqrt{1-4x}}{2x} \,\,\,\,\,\, \forall \vert x \vert < \dfrac14$$ This is the generating function for the Catalan numbers. Hence, in our case, we get that $$\sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}} = \dfrac{1-\sqrt{1-4 \cdot \left(\dfrac{b}{2a} \right)^2}}{2 \cdot \left(\dfrac{b}{2a} \right)^2} \,\,\,\,\,\,\,\,\forall \dfrac{b}{2a} < \dfrac12$$ Hence, $$\dfrac{aI}{\pi} = \dfrac{1-\sqrt{1-\left(\dfrac{b}a\right)^2}}{\dfrac{b^2}{2a^2}} = \dfrac{a-\sqrt{a^2-b^2}}{\dfrac{b^2}{2a}}$$ Hence, $$I = \dfrac{2\pi}{b^2} (a-\sqrt{a^2-b^2})$$

  • $\begingroup$ Catalan numbers?? yeah...it's new for me... :D thanks so much :)) $\endgroup$ – ochapuspita Mar 10 '13 at 20:52
  • $\begingroup$ Something like that i really didn't expect $\endgroup$ – Dominic Michaelis Mar 10 '13 at 20:53
  • $\begingroup$ what do you mean? $\endgroup$ – ochapuspita Mar 10 '13 at 20:56
  • $\begingroup$ @ochapuspita I assume that Dominic Michaelis's comment was on my answer and not on your previous comment. $\endgroup$ – user17762 Mar 10 '13 at 20:57

You can solve it with direct integration some subsitutions gives you the antiderivative: $$\frac{-2 \sqrt{b^2-a^2} \tanh ^{-1}\left(\frac{(a-b) \tan \left(\frac{\theta }{2}\right)}{\sqrt{b^2-a^2}}\right)+a \theta -b \sin (\theta )}{b^2}$$ use a substituion like $t=\tan(\theta)$. (and a lot of the trigonometric identies.)

We use the identies \begin{align*} \cos(\arctan(t))&=\frac{1}{\sqrt{1+t^2}}\\ \sin(\arctan(t))&=\frac{t}{\sqrt{1+t^2}} \end{align*}

So we have $$\int \frac{\sin^2(\theta)}{a+b\cos(\theta)}\, \mathrm{d}\theta= \int \frac{t^2}{a + b \frac{1}{\sqrt{1+t^2}}} \, \mathrm{d} t$$

With residue theorem it should work like this : $$\int_0^{2\pi} \frac{\sin^2(\theta)}{a+ b \cos(\theta)}\, \mathrm{d}\theta= -\int_0^{2\pi} \frac{1}{4}\cdot \frac{(\exp(i \theta)- \exp(-i \theta))^2}{a+\frac{b}{2} (\exp(i\theta) +\exp(-i\theta)} \, \mathrm{d}\theta$$

Chosing the way $\gamma(\theta) = e^{i\theta}$we see that it equal $$-\frac{1}{4i} \int_\gamma \frac{(z-z^{-1})^2}{a+ \frac{b}{2} (z+z^{-1})} \cdot \frac{1}{z} \, \mathrm{d} z$$ Expanding gives us $$-\frac{1}{4i} \int_\gamma \frac{z^2 -2 + \frac{1}{z^2}}{a+ \frac{b}{2} (z+\frac{1}{z})} \cdot \frac{1}{z} \, \mathrm{d}z=-\frac{1}{4i} \int_\gamma \frac{z^2 -2 + \frac{1}{z^2}}{za+ \frac{b}{2} (z^2+1)} \, \mathrm{d}z$$

  • $\begingroup$ $t=\tan\theta$ is enough (and simpler) once the part of the integral on $(\pi,2\pi)$ is brought back to $(0,\pi)$ by $\theta\to\theta-\pi$. $\endgroup$ – Did Mar 10 '13 at 19:16
  • $\begingroup$ @Did oh yes i will edit it $\endgroup$ – Dominic Michaelis Mar 10 '13 at 19:16
  • $\begingroup$ thank you so much....i'll try to undertand.... $\endgroup$ – ochapuspita Mar 10 '13 at 19:34
  • $\begingroup$ i always wrong in writing latex formula..:p $\endgroup$ – ochapuspita Mar 10 '13 at 20:27
  • $\begingroup$ $ \sin \theta =\frac{z-z^{-1}}{2i} $ $ \cos \theta =\frac{z-z^{-1}}{2} $ $\endgroup$ – ochapuspita Mar 10 '13 at 20:32

First step: $\cos(\theta+\pi)=-\cos(\theta)$ and $\sin^2(\theta+\pi)=\sin^2(\theta)$ hence the integral $I$ to be computed is $$ I=\int_0^\pi\frac{\sin^2\theta}{a+b\cos\theta}\mathrm d\theta+\int_0^\pi\frac{\sin^2\theta}{a-b\cos\theta}\mathrm d\theta=2aJ, $$ with $$ J=\int_0^\pi\frac{\sin^2\theta}{a^2-b^2\cos^2\theta}\mathrm d\theta. $$ Second step: the transformation $\theta\to\theta+\pi$ leaves the integrand unchanged hence one uses the change of variable $t=\tan\theta$, $\mathrm dt=(1+t^2)\mathrm d\theta$, which yields $$ J=\int_{-\infty}^{+\infty}\frac{t^2}{a^2(1+t^2)-b^2}\frac{\mathrm dt}{1+t^2}. $$ Third step: the fraction with argument $t^2$ can be decomposed as $$ \frac{x}{(a^2(1+x)-b^2)(1+x)}=\frac1{b^2}\left(\frac{1}{1+x}-\frac{a^2-b^2}{a^2(1+x)-b^2}\right), $$ hence $b^2J=K-(a^2-b^2)L$ with $$ K=\int_{-\infty}^{+\infty}\frac{\mathrm dt}{1+t^2},\qquad L=\int_{-\infty}^{+\infty}\frac{\mathrm dt}{a^2t^2+a^2-b^2}. $$ Fourth step: the change of variable $at=\sqrt{a^2-b^2}s$ yields $$ L=\frac{\sqrt{a^2-b^2}}a\int_{-\infty}^{+\infty}\frac{\mathrm ds}{(a^2-b^2)(1+s^2)}=\frac{K}{a\sqrt{a^2-b^2}}, $$ and the integral $K$ is classical, its value is $K=\pi$.

Conclusion: $$ I=2aJ=\frac{2a}{b^2}\left(K-(a^2-b^2)\frac{K}{a\sqrt{a^2-b^2}}\right)=\frac{2\pi}{b^2}\left(a-\sqrt{a^2-b^2}\right) $$


I'll do this one $$\int_{0}^{2\pi}\frac{cos(2\theta)}{a+bcos(\theta)}d\theta$$if we know how to do tis one you can replace $sin^{2}(\theta)$ by $\frac{1}{2}(1-cos(2\theta))$ and do the same thing.
if we ae on the unit circle we know that $cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ so by letting $z=e^{i\theta}$ we wil get$$cos(\theta)=\frac{z+\frac{1}{z}}{2}=\frac{z^2+1}{2z}$$and$$cos(2\theta)=\frac{z^2+\frac{1}{z^2}}{2}=\frac{z^4+1}{2z^2}$$ thus, if$\gamma: |z|=1$ the integral becomes $$\int_{\gamma}\frac{\frac{z^4+1}{2z^2}}{a+b\frac{z^2+1}{2z}}\frac{1}{iz}dz=\int_{\gamma}\frac{-i(z^4+1)}{2z^2(bz^2+2az+b)}dz$$now the roots of $bz^2+2az+b$ are $z=\frac{-2a\pm \sqrt{4a^2-4b^2}}{2b}=\frac{-a}{b}\pm\frac{\sqrt{a^2-b^2}}{b}$, you can check that the only root inside $|z|=1$ is $z_1=\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}$ so the only singularities of the function that we want to integrate inside $\gamma$ are $z_0=0$ and $z_1$ both are poles. find the resudies and sum them to get the answer.

Notice that this will only give youe "half" the answer you still have to do$$\frac{1}{2}\int_{\gamma}\frac{d\theta}{a+bcos(\theta)}$$ in the same way to get the full answer.

  • $\begingroup$ yeah, but i couldn't find $\frac{2\pi}{b^2}$ on the right. it's just $\frac{2\pi}{b}$. maybe there's some missing. i'll check again... $\endgroup$ – ochapuspita Mar 10 '13 at 20:50
  • $\begingroup$ thanks for sharing your answer... :) $\endgroup$ – ochapuspita Mar 10 '13 at 21:03

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