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Find value of integral: $$I_1=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$$ and $$I_2=\int_0^{2\pi}\frac{dx}{(2+\sin x)^2}$$

I don't know how, i need a solution, please

Iloveyou
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3 Answers3

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We can also use contour integration.

Let $z=e^{ix}$, then $\mathrm{d}x=-i\,\mathrm{d}z/z$ and $\cos(x)=\frac{z+1/z}{2}$ $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2} &=\oint\frac{-i\,\mathrm{d}{z}/z}{\left(2+\frac{z+1/z}{2}\right)^2}\\ &=\oint\frac{-4iz\,\mathrm{d}{z}}{(z^2+4z+1)^2}\tag{1} \end{align} $$ The zeros of $z^2+4z+1$ are $-2\pm\sqrt3$. $-2-\sqrt3$ is outside the unit circle, so the only pole that matters is at $-2+\sqrt3$.

Partial fractions gives $$ \frac{z}{(z-a)^2(z-b)^2}=\frac{\frac{a+b}{(b-a)^3}}{z-a}+\frac{\frac{b+a}{(a-b)^3}}{z-b}+\frac{\frac{a}{(a-b)^2}}{(z-a)^2}+\frac{\frac{b}{(b-a)^2}}{(z-b)^2}\tag{2} $$ With $a=-2+\sqrt3$ and $b=-2-\sqrt3$, we get the residue of $(2)$ at $z=a$ to be $\frac1{6\sqrt3}$. Plugging this into $(1)$ gives $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2} &=(2\pi i)(-4i)\frac1{6\sqrt3}\\ &=\frac{4\pi}{3\sqrt3}\tag{3} \end{align} $$

My original inclination was to use Igor's substitution, just to verify the previous answer, I will compute using the substitution $z=\tan(x/2)$, where $\mathrm{d}x=\frac{2\,\mathrm{d}z}{1+z^2}$ and $\cos(x)=\frac{1-z^2}{1+z^2}$ : $$ \begin{align} &\int_0^{2\pi}\frac{\mathrm{d}x}{(2+\cos(x))^2}\\ &=\int_{-\infty}^\infty\frac{\frac{2\,\mathrm{d}z}{1+z^2}}{\left(2+\frac{1-z^2}{1+z^2}\right)^2}\\ &=\int_{-\infty}^\infty\frac{2(1+z^2)\,\mathrm{d}z}{\left(3+z^2\right)^2}\\ &=\int_{-\infty}^\infty\frac{2\,\mathrm{d}z}{3+z^2}-\int_{-\infty}^\infty\frac{4\,\mathrm{d}z}{\left(3+z^2\right)^2}\\ &=\frac2{\sqrt3}\left[\tan^{-1}\left(\frac{z}{\sqrt3}\right)\right]_{-\infty}^\infty-\frac2{3\sqrt3}\left[\tan^{-1}\left(\frac{z}{\sqrt3}\right)+\frac{\sqrt3z}{3+z^2}\right]_{-\infty}^\infty\\ &=\frac{4\pi}{3\sqrt3}\tag{4} \end{align} $$

robjohn
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  • Very Nice solution...... To robjohn would you like to give me a document about contour integration , Thanks – juantheron Dec 12 '13 at 16:59
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Use the $u = \tan x/2$ substitution, which transforms this into a rational function.

Igor Rivin
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  • $u=\tan\frac{x}{2}$, when $2\pi$ then $\tan\frac{x}{2}$ is not identified – Iloveyou Dec 12 '13 at 16:00
  • if we let $u=\tan x/2$ it seems like the index are both zero,$\tan 2\pi/2=\tan 0/2=0$,the integral does not converge.... – Jonas Kgomo Dec 12 '13 at 16:05
  • the indefinite intergral is $\int \frac{2du}{(2+\frac{1-u^2}{1+u^2})^2(1+u^2)}$,taking $\cos x=\frac{1-u^2}{1+u^2},du=\frac{2du}{1+u^2}dx$,but the index is the problem,since with the indexes we have $\int _0^0f(x)dx$ which does not converge – Jonas Kgomo Dec 12 '13 at 16:16
  • So break it up into two integrals: from $0$ to $\pi$ and from $\pi$ to $2\pi.$ – Igor Rivin Dec 12 '13 at 17:04
  • Jeff is still asking how your work on $\rm DEFINT$ is coming along? ;-) – Bill Dubuque Dec 16 '13 at 00:31
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Let $\displaystyle I = \frac{\sin x}{(2+\cos x)}$

Now Diff. both side w.r. to $x$ , $\displaystyle \frac{dI}{dx} = \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{(2+\cos x)\cdot \cos x-\sin x\cdot (-\sin x)}{(2+\cos x)^2}$

$\displaystyle \frac{dI}{dx} = \frac{2\cos x+1}{(2+\cos x)^2}\Rightarrow \frac{dI}{dx} = \frac{2\cdot \left(2+\cos x\right)-3}{(2+\cos x)^2} = \frac{2}{2+\cos x}-3\cdot \frac{1}{(2+\cos x)^2}$

Now Integrate both side w. r. to $x$

$\displaystyle \int \frac{dI}{dx}dx = 2\int\frac{1}{(2+\cos x)}dx - 3\int\frac{1}{(2+\cos x)^2}dx$

So $\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{2}{3}\int\frac{1}{(2+\cos x)}dx-\frac{1}{3}\cdot I$

Now Put $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$

$\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{2}{3}\int\frac{1+\tan^2 \frac{x}{2}}{2+2\tan^2 \frac{x}{2}+1-\tan^2 \frac{x}{2}}dx-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$

$\displaystyle = \frac{2}{3}\int\frac{\sec^2 \frac{x}{2}}{3+\tan^2 \frac{x}{2}}dx-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$

Let $\displaystyle \tan \frac{x}{2} = t$ and $\sec^2\frac{x}{2}dx = 2dt$

$\displaystyle = \frac{4}{3}\int \frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$

$\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{4}{3}\cdot \frac{1}{\sqrt{3}}\cdot \tan^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{3}}\right)-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$

Yes Mrnhan $\displaystyle \tan \frac{x}{2}$ is not defined at $\displaystyle x = \pi$

Thanks Alraxite.

juantheron
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