TITLE: Can this integral $\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$ be written in the form of  a elliptic integral
QUESTION [14 upvotes]: I am trying to find the magnetic field due to an elliptic loop of wire. How to do integrals of the type $$\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$$ Where a and b are the axes of the ellipse. From dimensional considerations in the problem, I have deduced that the value of the integral must be must reduced to $$\frac{ L}{(ab)^2}$$ where $L$ is the circumference of the ellipse. How do I show it. 
OK. So now I know from wikipedia that the circumference of ellipse involves something called elliptic integrals, so thankfully, I wouldnt have to be looking for a closed form. But does this reduce to the elliptic integral? Or have I done something wrong?

REPLY [4 votes]: As a last post before leaving, I think I should add some comment on how I guessed the correct form of the integral. Basically, my motivation was to find the Magnetic field at the center of an elliptic loop, so I use physics notation and apologies for all the non-rigorous trivialities. 
Using  polar coordinates for the ellipse: $$r(\theta) = \frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$$ The actual integral I needed to evaluate was $$\frac{\mu_0 I }{4\pi}\oint \frac{\vec{dl}\times\hat{r}}{r^2}$$ (Notation:$\vec{dl}$ is a "small" vector following the elliptic loop. $r,\hat{r},\vec{r}$ are magnitude, direction, and vectors from the origin to $\vec{dl}$)
Using the fact that $$d\theta = \frac{|\vec{dl}\times\hat{r}|}{r}$$ I get $$\frac{\mu_0 I\hat{k} }{4\pi ab}\int_0^{2\pi}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}d\theta$$ (EDIT: A look at the integrand immediately tells you that this is the circumference of an ellipse with semidiameters $a$ and $b$. What follows is formal.)
Now let $$1-\frac{b^2}{a^2} = k^2$$ we have $$\frac{\mu_0I\hat{k}}{\pi b}\int_0^{\pi/2}\sqrt{1-k^2 \sin^2\theta}d\theta$$ $$=\frac{\mu_0 I L}{4A}$$ L is circumference and A is area.
So it is clear that its much simple in polar form, but I had tried to do it in the cartesian form initially so I was struggling to show how the two were equivalent. In a larger context, there might be something connecting the two integrals, but I'll leave that to others.