<s>[INST] 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 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?
 [/INST] (I assume that $a > b$ in this answer.)
Hans's answer covers the OP's needs, but I'll try out the pedestrian approach here.
First, a trivial simplification:
$$\int_0^{2\pi} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}=\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}+\int_\pi^{2\pi} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
$$=\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}+\int_0^{\pi} \frac{\mathrm d\theta}{(a^2 \cos^2(\theta+\pi)+b^2\sin^2(\theta+\pi))^{3/2}}$$
$$=2\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
Another simplification can be done as follows:
$$=2\int_{-\pi/2}^{\pi/2} \frac{\mathrm d\theta}{(a^2 \cos^2(\theta+\pi/2)+b^2\sin^2(\theta+\pi/2))^{3/2}}$$
$$=2\int_{-\pi/2}^{\pi/2} \frac{\mathrm d\theta}{(a^2 \sin^2\theta +b^2\cos^2\theta)^{3/2}}$$
or, since the integrand is an even function,
$$=4\int_{-\pi/2}^0 \frac{\mathrm d\theta}{(a^2 \sin^2\theta +b^2\cos^2\theta)^{3/2}}$$
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
Now, to get the integral into something recognizable, we use the Pythagorean identity and then try to factor out constants, like so:
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2(1-\sin^2\theta)+b^2\sin^2\theta)^{3/2}}$$
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2-(a^2-b^2)\sin^2\theta)^{3/2}}$$
$$=\frac4{a^3}\int_0^{\pi/2} \frac{\mathrm d\theta}{(1-m\sin^2\theta)^{3/2}}$$
where we let $m=1-\frac{b^2}{a^2}$, and then introduce the substitution $\theta=\mathrm{am}(u|m)$, $\mathrm d\theta=\mathrm{dn}(u|m)\mathrm du$ (where $\mathrm{am}(u|m)$ is the Jacobian amplitude and $\mathrm{dn}(u|m)$ is a Jacobian elliptic function):
$$=\frac4{a^3}\int_0^{K(m)} \frac{\mathrm{dn}(u|m)\mathrm du}{(1-m\mathrm{sn}^2(u|m))^{3/2}}$$
(since $\sin(\mathrm{am}(u|m))=\mathrm{sn}(u|m)$). Using the identity $\mathrm{dn}^2(u|m)+m\mathrm{sn}^2(u|m)=1$, the integral turns into
$$=\frac4{a^3}\int_0^{K(m)} \frac{\mathrm du}{\mathrm{dn}^2(u|m)}=\frac4{a^3}\int_0^{K(m)} \mathrm{nd}^2(u|m)\mathrm du$$
Using formula 22.16.20 in the DLMF, the integral evaluates to
$$\frac4{a^3}\frac{E(m)}{1-m}=\frac4{ab^2}E\left(1-\frac{b^2}{a^2}\right)$$
where $E(m)$ is the complete elliptic integral of the second kind, and since $L=4aE\left(1-\frac{b^2}{a^2}\right)$, your supposition is correct.
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