Elliptic integrals Knowing that 
$$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}t+b^2\sin^{2}t}dt＝\int_{0}^{2\pi}\sqrt[]{1+\cos^{2}t}dt$$
how to find the value of $a$ and $b$?
 Elliptic integral? Thank you!
 A: $$\int^{2\pi}_{0}\sqrt{a^2\cos^{2}(t)+b^2\sin^{2}(t)}dt = |a|\int^{2\pi}_{0}\sqrt{1-\left(1-\frac{b^2}{a^2}\right)\sin^{2}(t)}dt =4|a|\text{E}\left(1-\frac{b^2}{a^2}\right)$$
E$(X)$ is the complete elliptic integral of second kind.
$$\int^{2\pi}_{0}\sqrt{1+\cos^{2}(t)}dt =\sqrt{2}\int^{2\pi}_{0}\sqrt{1-\frac{1}{2}\sin^{2}(t)}dt = 4\sqrt{2}\text{  E}\left(\frac{1}{2}\right)$$ 
The equation to be solved is :
$$|a|\text{E}\left(1-\frac{b^2}{a^2}\right) = \sqrt{2}\text{  E}\left(\frac{1}{2}\right)$$
Let $X=1-\frac{b^2}{a^2}\quad$ Hense $\quad X\leq 1\quad$ and E$(X)$ is real.
$a=\pm\frac{b}{\sqrt{1-X}}\quad$ This transforms the equation to :
$$\text{E}(X) = \sqrt{2}\text{  E}\left(\frac{1}{2}\right)\frac{\sqrt{1-X} }{|b|} \qquad (1)$$
On $-\infty<X\leq 1$ the functions E$(X)$ and $\sqrt{1-X}$ are both smooth functions. Both are smooth and decreassing.

Given a value of $b$, the curves E$(X)$ and $\sqrt{1-X}$ intersect each other on one point only. 
For example, given $b=1\quad\to\quad X=\frac{1}{2}\quad\to\quad a=\pm \sqrt{2}$
So, they are an infinity of solutions : To each arbitrary value of $b$, the transcendantal equation (1) has a root to which corresponds a value of $X$ and a value of $|a|$.
A: Note that $$\sqrt{1+\cos^2t}=\sqrt{2\cos^2t+\sin^2t}$$
A: Let $F(a,b) = \int^{2\pi}_{0}\sqrt{a^2\cos^{2}t+b^2\sin^{2}t}dt$ for $a, \, b \ge 0$. As is noted in the previous answer, $\sqrt{1 + \cos^2 t} = F(\sqrt{2},1)$.   
You are therefore asking whether the solution of $F(a,b) = F(\sqrt{2},1)$ is unique. Clearly that is not the case. The solution set is an analytic curve since $F$ is analytic and $\nabla F$ never vanishes. To obtain the parametrization, note that $F(a,b) = bF\left(\frac{a}{b},1 \right)$. This can then be worked out in terms of elliptic integrals of the second kind. 
