<s>[INST] The asymptotic behavior of an integral The integral in hand is
$$
I(n) = \frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}\, dx
$$
I dont know whether it has closed-form or not, but currently I only want to know its asymptotic behavior. Setting $x=\cos\theta$, then
$$
I(n) = \frac{1}{\pi}\int_{0}^{\pi/2} \Big[(1+2\cos\theta)^{2n}+(1-2\cos\theta)^{2n}\Big]\, d\theta
$$
The second term can be neglected, therefore
$$
I(n) \sim \frac{1}{\pi}\int_{0}^{\pi/2}(1+2\cos\theta)^{2n}\, d\theta
$$
How can I move on?
 [/INST] To compute the asymptotic expansion of the integral, we split it into two pieces
$$\frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx
= \frac{1}{\pi}\left(\int_{-1}^0 + \int_0^1\right)\frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx
$$
Over the interval $[-1,0]$, we have $|1+2x|\le 1$, so the contribution there is bounded.
$$\mathcal{I}_1 \stackrel{def}{=} \left|\frac{1}{\pi}\int_{-1}^0 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx\right| \le 
\frac{1}{\pi}\int_{-1}^0 \frac{dx}{\sqrt{1-x^2}} = \frac12
$$
Over the interval $[0,1]$, introduce variable 
$$1 + 2x = 3 e^{-t} \quad\iff\quad x = \frac{3e^{-t}-1}{2}$$
We have
$$\mathcal{I}_2 \stackrel{def}{=} \frac{1}{\pi}\int_{0}^1 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx = 
\frac{3^{2n+1}}{2\pi}\int_{0}^{\log 3} e^{-(2n+1)t} \left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2} dt$$
Notice near $t = 0$, the complicated mess of the square bracket has following Taylor series expansion:
$$\left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2}
= \frac{1}{\sqrt{3t}}\left(
1+ \frac{5}{8}t+ \frac{49}{384}t^2-\frac{29}{3072}t^3 + \cdots
\right)\tag{*1}$$
The whole expression is in the form which we can apply Watson's Lemma and read off the asympotic expansion:
$$\begin{align}
\mathcal{I}_2 \;\approx &\;
\frac{3^{2n+1}}{2\sqrt{3}\pi}
\left(
\frac{\Gamma\left(\frac12\right)}{\sqrt{2n+1}}
+ \frac{5}{8}\frac{\Gamma\left(\frac32\right)}{\sqrt{2n+1}^3}
+ \frac{49}{384}\frac{\Gamma\left(\frac52\right)}{\sqrt{2n+1}^5}
- \frac{29}{3072}\frac{\Gamma\left(\frac72\right)}{\sqrt{2n+1}^7}
+ \cdots
\right)\\
\;\approx &\;
\frac{3^{2n}\sqrt{3}}{2\sqrt{\pi(2n+1)}}
\left( 1 + \frac{5}{16(2n+1)} + \frac{49}{512(2n+1)^2}
 - \frac{145}{8192(2n+1)^3} + \cdots\right)
\end{align}
$$
Since $\mathcal{I}_1$ is always of $O(1)$, above asymbolic expansion for $\mathcal{I}_2$ is also the one for $\mathcal{I}_1 + \mathcal{I}_2$. i.e. the one you are looking for.
If one want more terms for the asymptotic expansion, one just need to throw the LHS of $(*1)$ to an CAS, crank out more terms of the Taylor expansion and repeat above process.
</s>