1. An integration problem

Here is the problem.
$\displaystyle \begin{displaymath} \int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta \end{displaymath}$

2. Easy way

Hi , solve by this way :

$\displaystyle \int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}}$

now you can use the this substitute :

$\displaystyle sin\theta = (a-cos\theta)sinx$

3. Originally Posted by parkhid
Hi , solve by this way :

$\displaystyle \int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}}$

now you can use the this substitute :

$\displaystyle sin\theta = (a-cos\theta)sinx$
But $\displaystyle (a-\cos\theta)^2-\sin^2\theta} \ne a^2-2a\cos{\theta}+1$, unless I'm missing something.

4. Originally Posted by wslgx1024
Here is the problem.
$\displaystyle \begin{displaymath} \int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta \end{displaymath}$
Where has the integral come from?

Read this: Elliptic Integral of the First Kind -- from Wolfram MathWorld

5. Originally Posted by mr fantastic
Where has the integral come from?

Read this: Elliptic Integral of the First Kind -- from Wolfram MathWorld
The original problem is

In electrostatics, a charge distribution over a curve $\displaystyle \mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $\displaystyle$\textbf{R}^{3}\backslash\mathcal{C}$.$
The charge distribution is given by the charge density $\displaystyle$\rho: \mathcal{C} \rightarrow \textbf{R}$$, and the induced electric potential at \displaystyle p \in \textbf{R}^{3}\backslash\mathcal{C}  is given by \displaystyle \begin{center} V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}} \end{center}  where \displaystyle \mathrm{dist(p,\cdot)} = \|p-x\|$$ denotes the distance between $\displaystyle$p\in \textbf{R}^{3}\backslash\mathcal{C}$$and \displaystyle x\in \mathcal{C}$$.
Consider a uniform charge density $\displaystyle$\rho$$on the set \displaystyle \begin{center} \mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}. \end{center}  Find the potential at points on the x-axis, |x|>1. And by using the parametrization \displaystyle x=\cos(\theta)$$ and $\displaystyle$y=\sin(\theta)$I got the previous integral. And how could the integral in the link you gave solve this integral? 6. Originally Posted by wslgx1024 The original problem is In electrostatics, a charge distribution over a curve$\displaystyle \mathcal{C}\subset \textbf{R}^{3}$induces an electric potential in$\displaystyle $\textbf{R}^{3}\backslash\mathcal{C}$.
$The charge distribution is given by the charge density$\displaystyle $\rho: \mathcal{C} \rightarrow \textbf{R}$$, and the induced electric potential at \displaystyle p \in \textbf{R}^{3}\backslash\mathcal{C} is given by \displaystyle \begin{center} V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}} \end{center} where \displaystyle \mathrm{dist(p,\cdot)} = \|p-x\|$$ denotes the distance between$\displaystyle $p\in \textbf{R}^{3}\backslash\mathcal{C}$$and \displaystyle x\in \mathcal{C}$$. Consider a uniform charge density$\displaystyle $\rho$$on the set \displaystyle \begin{center} \mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}. \end{center} Find the potential at points on the x-axis, |x|>1. And by using the parametrization \displaystyle x=\cos(\theta)$$ and$\displaystyle $y=\sin(\theta)$

I got the previous integral.
And how could the integral in the link you gave solve this integral?
I thought that might be the case. These sorts of integrals often occur in those contexts. You need to re-arrange your integral to recognise the elliptic-integral form you have. Evaluate your integral here: http://www.wolframalpha.com/ to see where you need to head.

7. Originally Posted by mr fantastic
I thought that might be the case. These sorts of integrals often occur in those contexts. You need to re-arrange your integral to recognise the elliptic-integral form you have. Evaluate your integral here: Wolfram|Alpha&mdash;Computational Knowledge Engine to see where you need to head.
Thanks a lot for help. BTW, the website is great!