Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.

$$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$

share|improve this question
    
Are you familiar with methods of complex contour integration? (residue theorem). It should be fairly straight-forward. –  orion May 26 at 15:46
    
no i am not familiar with complex analytical methods –  rockstar123 May 26 at 15:59

6 Answers 6

Consider the following integral: $$I(a)=\int_0^{\pi} \frac{d\theta}{a+\cos\theta}$$ Rewrite $\cos\theta=\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$ to get: $$I(a)=\int_0^{\pi} \frac{\sec^2(\theta/2)}{a+1+\tan^2(\theta/2)(a-1)}d\theta$$ Use the substitution $\tan(\theta/2)=t \Rightarrow \sec^2(\theta/2)d\theta=2\,dt$ to get: $$I(a)=\int_0^{\infty} \frac{2}{a+1+(a-1)t^2}\,dt$$ The above is easy to evaluate and gives: $$I(a)=\frac{\pi}{\sqrt{a^2-1}}$$ Hence, $$\int_0^{\pi} \frac{d\theta}{a+\cos\theta}=\frac{\pi}{\sqrt{a^2-1}}$$ Differentiate both the sides wrt $a$ to get: $$\int_0^{\pi} \frac{d\theta}{(a+\cos\theta)^2}=\frac{a\pi}{(a^2-1)^{3/2}}\,\,\,\,\,(*)$$ Getting back to the problem, you have, $$\int_0^{2\pi}\frac{d\theta}{(2+\cos\theta)^2}=2\int_0^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$$ From $(*)$ and with $a=2$, the final answer is: $$2\frac{2\pi}{(3\sqrt{3})}=\boxed{\dfrac{4\pi}{3\sqrt{3}}}$$

share|improve this answer
    
+1 for rewriting $\cos$ as a quotient of $\tan$'s. –  iHubble May 26 at 16:10
    
Thank you iHubble! :) –  Pranav Arora May 26 at 16:11

Here's the contour integration approach.

$$\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta = \int_{0}^{2 \pi} \frac{1}{(2+\frac{e^{i \theta} + e^{- i \theta}}{2})^{2}} \ d \theta$$

Let $z=e^{i \theta}$.

Then

$$ \begin{align}\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta &= \int_{|z|=1} \frac{1}{(2+\frac{z+z^{-1}}{2})^{2}} \frac{dz}{iz} \\ &= \frac{4}{i}\int_{|z|=1} \frac{z}{(z^{2}+4z+1)^{2}} \ d z \\ &=\frac{4}{i} \int_{|z|=1}\frac{z}{[(z-\sqrt{3}+2)(z+\sqrt{3}+2)]^{2}} \ dz \end{align}$$

The only pole inside the unit circle is at $z= \sqrt{3}-2$.

Therefore,

$$ \begin{align} \int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta &= 2 \pi i \frac{4}{i} \text{Res} \left[ \frac{z}{[(z-\sqrt{3}+2)(z+\sqrt{3}+2)]^{2}}, \sqrt{3}-2 \right] \\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{d}{dz} \frac{z}{(z+\sqrt{3}+2)^{2}} \\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{(z+\sqrt{3}+2)^{2}-2z(z+\sqrt{3}+2)}{(z+\sqrt{3}+2)^{4}} \\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{-z+ \sqrt{3} + 2}{(z+\sqrt{3}+2)^{3}} \\ &= 8 \pi \frac{4}{24 \sqrt{3}} = \frac{4 \pi}{3 \sqrt{3}} \end{align}$$

share|improve this answer
    
Nice Solution....... –  juantheron May 31 at 3:26
    
@juantheron Thanks. –  Random Variable May 31 at 3:34

Here is an ugly way of doing it. Pranav's solution is much more elegant.

First substitute $u = \tan(\tfrac{\theta}{2})$ and $\mathrm{d}u = \tfrac{1}{2} \sec^2(\tfrac{\theta}{2})\mathrm{d}\theta$. Then, transform the integrand using the substitution $\sin(\theta) = \frac{2u}{u^2+1}$, $\cos(\theta) = \frac{1-u^2}{u^2+1}$ and $\mathrm{d}\theta = \frac{2\mathrm{d}u}{u^2+1}$ to get $$I = \int \frac{1}{(2+\cos(\theta))^2}\mathrm{d}\theta = \int \frac{2}{(u^2 + 1)\left(\tfrac{1-u^2}{u^2+1} +2\right)^2} \mathrm{d}u = 2\int \frac{u^2+1}{u^4 + 6u^2 +9} \mathrm{d}u.$$

Now, using partial fractions, one can rewrite this integral as $$I = \frac{2}{3}\int\frac{1}{\tfrac{u^2}{3}+1}\mathrm{d}u - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$

For the first integrand, substitute $s = \frac{u}{\sqrt{3}}$ and $\mathrm{d}s = \frac{1}{\sqrt{3}}\mathrm{d}u$ to get $$I = \frac{2}{\sqrt{3}} \int \frac{1}{s^2+1} \mathrm{d}s - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$ We know how to integrate the left side. This is just $$I = \frac{2\arctan(s)}{\sqrt{3}} - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$

For the right-hand side, we substitute $u = \sqrt{3}\tan(p)$ and $\mathrm{d}u=\sqrt{3}\sec^2(p)\mathrm{d}p$. Then, we have that $(u^2+3)^2 = (3\tan^2(p) +2)^2 = 9\sec^4(p)$ and $p = \arctan(\tfrac{u}{\sqrt{3}})$. This yields the integral $$I = \frac{2\arctan(s)}{\sqrt{3}} - \frac{4}{3\sqrt{3}} \int \cos^2(p) \mathrm{d}p.$$ Integrating $\cos^2(p)$ is easy by using the identity $\cos^2(p) = \frac{\cos(2p) + 1}{2}$. All that remains now is to integrate this and substitute your way back to express $I$ in terms of $\theta$. Evaluate from $0$ to $2\pi$ to obtain $$I_0^{2\pi} = \frac{4\pi}{3\sqrt{3}}.$$

share|improve this answer
3  
We have same approach bro. :D –  Tunk-Fey May 26 at 16:06
1  
In all honesty, we have Wolfram's approach. ;) –  iHubble May 26 at 16:07

HINT :

Using Weierstrass substitution, the integral turns out to be $$ \int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta=2\int_0^{\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta=4\int_0^\infty\frac{1+x^2}{(3+x^2)^2}\ dt. $$

Now, using partial fraction decomposition, the integrand turns out to be $$ \frac{1+x^2}{(3+x^2)^2}=\frac{1}{3+x^2}-\frac{2}{(3+x^2)^2}. $$ The last step, using substitution $x=\sqrt3\tan t$, the integral should be easy to be evaluated.

share|improve this answer
    
The term "Weierstrass substitution" is a misnomer perpetuated by the popularity of Stewart's calculus textbooks (which is probably not the earliest appearance of the term). –  Michael Hardy Sep 6 at 16:08
    
@MichaelHardy Indeed, I know that. You've mentioned it many times here on MSE. Do you prefer tangent-half substitution to Weierstrass substitution? $\ddot\smile$ –  Tunk-Fey Sep 6 at 16:32
    
I do prefer to call it that. History is not mathematics, but still it's not good to confuse people about it. –  Michael Hardy Sep 6 at 16:47
    
@MichaelHardy OK, as you wish Sir, I'll use the term tangent-half sub from now on. –  Tunk-Fey Sep 6 at 16:58

Of course you can apply the substitution $t=\tan \frac{\theta}{2}$ directly. $$d\theta =\frac{2dt}{1+t^2}$$ and $$\cos \theta =\frac{1-t^2}{1+t^2}$$ So $$2\int_0^{\pi}\frac{d \theta}{(2+\cos\theta)^2}=4\int_0^{\infty} \frac{t^2+1}{(t^2+3)^2}dt$$

share|improve this answer

Google the term "tangent half-angle substitution".

If $u = \tan\dfrac\theta 2$ then $\cos\theta=\dfrac{1-u^2}{1+u^2}$ and $d\theta = \dfrac{2\,du}{1+u^2}$.

As $\theta$ goes from $0$ to $2\pi$, then $u$ goes from $-\infty$ to $\infty$.

If you remember some trigonometry, you can derive what you see above.

\begin{align} & \int_0^{2\pi}\frac{d\theta}{(2+\cos\theta)^2} = \int_{-\infty}^\infty \frac{\left(\frac{2\,du}{1+u^2}\right)}{\left(2+\frac{1-u^2}{1+u^2}\right)^2} = \int_{-\infty}^\infty \frac{2(1+u^2)\,du}{(2(1+u^2)+(1-u^2))^2} \\[10pt] = {} & \frac 1 2 \int_{-\infty}^\infty \frac{1+u^2}{(3+u^2)^2} \, du \end{align}

There still a lot of work to do after that. Use partial fractions. If your difficulty was in the part that comes after that, this won't help; otherwise maybe it will.

Perhaps this is an argument for using complex variables; residues, etc.: pursuing what you see above is in some ways routine but is onerous.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.