Arithmetic-Geometric limit $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n$ 
For $x,y>0$, define two sequences $(x_n)$ and $(y_n)$ by $x_1=x,y_1=y$ and $x_{n+1}=(x_n+y_n)/2$ and $y_{n+1}=\sqrt{x_ny_n}$. Prove that $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n= \dfrac{\pi}{\int_0^\pi \dfrac{d\theta}{\sqrt{x^2 \cos^2\theta + y^2\sin^2\theta}}}.$

I think it might be easier to prove $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n.$ Let the LHS of this equation be denoted $L$ and let the RHS be denoted $M$. By the AM-GM inequality and induction, $x_{n}\leq y_n$ for all $n\ge 2$. It could be useful to define a new sequence with a limit that's easier to evaluate. We have $L\leq M$ by limit properties, so we just need to show $L\ge M$ to get $L=M$. Suppose for a contradiction that $L < M.$ Then by definition, there exists $N$ so that for all $n\ge N, x_n < \frac{L+M}2$ and $y_n > \dfrac{L+M}2.$ How can I proceed from here?
As for showing it equals an expression involving a given integral, I think the integral is actually fairly hard to compute explicitly, so one should use some properties of the sequences to show the desired equality.

Putting together (essentially) the proof from comments that the limits exist and are equal, by @AbhijeetVats and @DanielWainfleet:
If $x=y$, then $x_n=y_n=x=y$ for all $n$, so obviously the limits exist and are equal.
If $x \neq y$, then $y_2 < x_2$ by the AM-GM inequality. Also note $y_2 > 0$. Now prove by induction that for every $n \geq 2$,
$$ y_2 \leq y_n < y_{n+1} < x_{n+1} < x_n \leq x_2 $$
Since $x_n > y_n > 0$, $y_{n+1} = \sqrt{x_n y_n} > \sqrt{y_n y_n} = y_n$. Since $y_n \geq y_2$, this also shows $y_{n+1} > y_2$.
Since $y_n < x_n$, $x_{n+1} = \frac{x_n+y_n}{2} < \frac{x_n+x_n}{2} = x_n$.
And by the AM-GM inequality again, $y_{n+1} < x_{n+1}$. The induction proof is complete.
So $(x_n)$ is a strictly decreasing sequence bounded below by $y_2$, and therefore converges to a real value $L = \lim_{n \to \infty} x_n$. And $(y_n)$ is a strictly increasing sequence bounded above by $x_2$, and therefore converges to a real value $M = \lim_{n \to \infty} y_n$.
But then
$$ L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac{x_n+y_n}{2} = \frac{M+L}{2} $$
which implies that $L=M$:
$$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$$
