Limit of integral of sequence. Sequence is not dominated by L1 function I'd like to prove
$$\lim_{n\rightarrow+\infty} \int_0^\pi n\sqrt{n^2+x^2-2nx\cos\theta}-n^2 \mathrm{d}\theta = \frac{\pi}{4}x^2$$
for $x\in[0,\infty)$. I checked it numerically and derived it with Matlab Symbolic Toolbox, but cannot prove it by calculus.
I cannot use the Dominated Convergence Theorem since I could not find any $L^1[0,\pi]$ function of $\theta$ dominating $n\sqrt{n^2+x^2-2nx\cos\theta}-n^2$. I also tried integration by parts, but it didn't work.
Has anyone any idea? (change of variable to get rid of $\cos\theta$, perhaps?)
Thanks 

Update, following Claude's solution
I'm trying to follow Claude's solution, but I've encountered a problem.
I had tried before to transform my integral into the complete elliptic integral of the second kind,
Following your notation,
\begin{equation*}
\begin{array}{rcl}
\displaystyle{\int_0^\pi \sqrt{1-k\cos\theta}~d\theta} &=&
\displaystyle{\int_0^\pi \sqrt{1-k\left(1-2\sin^2\frac{\theta}{2}\right)}}~d\theta \\
&=& \displaystyle{\int_0^\frac{\pi}{2} \sqrt{1-k+2k\sin^2\theta}}~d\theta \\
&=& 2 \sqrt{1-k}\displaystyle{\int_0^\frac{\pi}{2}}\sqrt{1+\frac{2k}{1-k}\sin^2\theta}~d\theta
\end{array}
\end{equation*}
But then I got stuck, because the complete elliptic integral of the second kind is of the form
$$\int_0^\frac{\pi}{2} \sqrt{1-a^2\sin^2\theta}~d\theta$$
with $a\in[0,1]$, right? Where did you get your 2nd equation from?
 A: It seems that you are entering the world of elliptic integrals.
$$I=\int\sqrt{n^2+x^2-2nx\cos\theta} \,d\theta = \sqrt{n^2+x^2} \int \sqrt{1- \frac {2nx}{n^2+x^2}\cos\theta}\,d\theta $$ and $$\int \sqrt{1-k\cos\theta}\,d\theta=-2 \sqrt{k-1} E\left(\frac{\theta}{2}|\frac{2 k}{k-1}\right)$$ making the definite integral 
$$ J=\int_0^\pi n\sqrt{n^2+x^2-2nx\cos\theta}-n^2 \,d\theta = 2 n (n+x)\,E\left(\frac{4 n x}{(n+x)^2}\right)-\pi  n^2$$ Now, using the expansion around $t=0$$$E(t)=\frac{\pi }{2}-\frac{\pi  t}{8}-\frac{3 \pi  t^2}{128}+O\left(t^3\right)$$ will show the limit and how it is approached.
$$J=\frac{\pi  x^2}{4}+\frac{\pi  x^4}{64 n^2}+O\left(\frac{1}{n^3}\right)$$
A: The integrand equals
$$\tag 1 n^2(1+x^2/n^2 -(2x\cos \theta)/n)^{1/2} - n^2.$$
From Taylor we have $(1+u)^{1/2} = 1+u/2 -u^2/8 +O(u^3)$ as $u\to 0.$ Apply this with $u = x^2/n^2 -(2x\cos \theta)/n$ to see
$(1)$ equals
$$\frac{n^2}{2} (x^2/n^2 -(2x\cos \theta)/n) - \frac{n^2}{8}\frac{4x^2\cos^2 \theta}{n^2} + O(\frac{1}{n}).$$
If you now integrate on $[0,\pi],$ you get $x^2\pi/4 + O(1/n) \to x^2\pi/4.$
