how to compute $\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n$? I need to compute $$\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n dx.$$
I proved that
for $n\ge1$,
$$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!},$$
but I don't know how to continue from here.
I also need to calculate $\int_{0}^{1}(1-x^2)^ndx$ for $n=50$ with a $1$% accuracy. I thought about using Taylor series but also failed.
 A: Let $y = \sqrt{n}x$. Then we have that
$$\lim_{n\to\infty} \int_0^{\sqrt{n}} \left(1-\frac{y^2}{n}\right)^n\:dy \longrightarrow \int_0^\infty e^{-y^2}\:dy = \frac{\sqrt{\pi}}{2}$$
by dominated convergence.

$\textbf{EDIT}$: To get the numerical accuracy you desire, we can make use of the following result describing the rate of convergence of the limit:
$$\left(1+\frac{x}{n}\right)^n = e^x - \frac{x^2e^x}{2n} + O\left(\frac{1}{n^2}\right)$$
To only achieve $1\%$ accuracy we can take a few liberties with the calculation, for example, after using the substitution above:
$$\frac{1}{\sqrt{n}}\int_0^{\sqrt{n}} \left(1-\frac{y^2}{n}\right)^n\:dy = \frac{1}{\sqrt{n}}\int_0^{\sqrt{n}} e^{-y^2}-\frac{1}{2n}y^4e^{-y^2}\:dy + O\left(n^{-\frac{5}{2}}\right)$$
$$ \approx \frac{1}{\sqrt{n}}\int_0^{\infty} e^{-y^2}\:dy - \frac{1}{2\sqrt{n^3}}\int_0^{\infty}y^4e^{-y^2}\:dy$$
where we can throw out the $O\left(n^{-\frac{5}{2}}\right)$ terms and approximate the integrals to $\infty$ since for $\sqrt{n} = 5\sqrt{2}$ the integrals have accumulated most of their area up to several decimal places.
The value of the second integral is given by Feynman's trick
$$\int_0^\infty y^4 e^{-y^2}\:dy = \Biggr[\frac{d^2}{da^2}\int_0^\infty e^{-ay^2}\:dy \Biggr]_{a=1} = \frac{1}{2}\Biggr[\frac{d^2}{da^2} \sqrt{\frac{\pi}{a}} \Biggr]_{a=1} = \frac{3\sqrt{\pi}}{8}$$
This gives us a nice, tidy approximation of 
$$\int_0^1 (1-x^2)^{50}\:dx \approx \frac{397}{4000}\sqrt{\frac{\pi}{2}} \sim 0.12439$$
Compare this to the actual value
$$\int_0^1 (1-x^2)dx \sim 0.12440$$
and the zeroth order approximation
$$\frac{1}{10}\sqrt{\frac{\pi}{2}} \sim 0.12533$$
A: Answer to your second question. Since
$$
\frac{{\frac{{(2n)!!}}{{(2n + 1)!!}}}}{{\frac{{(2n + 2)!!}}{{(2n + 3)!!}}}} = \frac{{(2n)!!(2n + 3)!!}}{{(2n + 1)!!(2n + 2)!!}} = \frac{{n + \frac{3}{2}}}{{n + 1}}
$$
and
$$
\frac{{n + 2}}{{n + 1}} < \left( {\frac{{n + \frac{3}{2}}}{{n + 1}}} \right)^2  < \frac{{n + 1}}{n}
$$
for all $n\geq 1$, the sequence
$$
\sqrt {n + 1} \frac{{(2n)!!}}{{(2n + 1)!!}}
$$
is decreasing and the sequence
$$
\sqrt n \frac{{(2n)!!}}{{(2n + 1)!!}}
$$
is increasing. They both converge to the same limit which is (by Ninad Munshi's answer) $\sqrt{\pi}/2$. Hence,
$$
\frac{{\sqrt \pi  }}{2}\frac{1}{{\sqrt {n + 1} }} < \frac{{(2n)!!}}{{(2n + 1)!!}} = \int_0^1 {(1 - x^2 )^n dx}  < \frac{{\sqrt \pi  }}{2}\frac{1}{{\sqrt n }}.
$$
You can use this to show that
$$
0.124096 < \int_0^1 {(1 - x^2 )^{50} dx}  < 0.125332.
$$
A: You won't need a close form for the integral. Here is an easy way to do it :
Denoting $ \left(\forall n\in\mathbb{N}\right),\ W_{n}=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} : $
We have : \begin{aligned} \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{x}\sin^{n}{x}\,\mathrm{d}x} \\ &=\left[-\cos{x}\sin^{n}{x}\right]_{0}^{\frac{\pi}{2}}+n\displaystyle\int_{0}^{\frac{\pi}{2}}{\cos^{2}{x}\sin^{n-1}{x}\,\mathrm{d}x}\\ &=n\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(1-\sin^{2}{x}\right)\sin^{n-1}{x}\,\mathrm{d}x}\\ \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=n\left(W_{n-1}-W_{n+1}\right)\\ \iff \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=\displaystyle\frac{n}{n+1}W_{n-1} \end{aligned}
And since $ \left(W_{n}\right)_{n\in\mathbb{N}} $ is positive and decreasing, we have that : $$ \left(\forall n\geq 2\right),\ W_{n+1}\leq W_{n}\leq W_{n-1}\iff \displaystyle\frac{n}{n+1}\leq\displaystyle\frac{W_{n}}{W_{n-1}}\leq 1 $$
Thus $ \displaystyle\lim_{n\to +\infty}{\displaystyle\frac{W_{n}}{W_{n-1}}}=1 \cdot $
We can easily verify that the sequence $ \left(y_{n}\right)_{n\in\mathbb{N}} $ defined as following $ \left(\forall n\in\mathbb{N}\right),\ y_{n}=\left(n+1\right)W_{n}W_{n+1} $ is a constant sequence. (Using the recurrence relation that we got from the integration by parts to express $ W_{n+1} $ in terms of $ W_{n-1} $ will solve the problem)
Hence $ \left(\forall n\in\mathbb{N}\right),\ y_{n}=y_{0}=W_{0}W_{1}=\displaystyle\frac{\pi}{2} \cdot $
Now that we've got all the necessary tools, we can prove that $ \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}}=\sqrt{\displaystyle\frac{\pi}{2}} : $ \begin{aligned} \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}} &=\displaystyle\lim_{n\to +\infty}{\sqrt{y_{n-1}}\sqrt{\displaystyle\frac{W_{n}}{W_{n-1}}}}\\ &=\displaystyle\lim_{n\to +\infty}{\sqrt{\displaystyle\frac{\pi}{2}}\sqrt{\displaystyle\frac{W_{n}}{W_{n-1}}}}\\ \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}}&=\sqrt{\displaystyle\frac{\pi}{2}} \end{aligned}
Using the substitution $ \left\lbrace\begin{aligned}x&=\cos{y}\\ \mathrm{d}x&=-\sin{y}\,\mathrm{d}y\end{aligned}\right. $, we can see that : $$ \left(\forall n\in\mathbb{N}\right),\ \int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin^{2n+1}{y}\,\mathrm{d}y}=W_{2n+1} $$
Thus $$ \lim_{n\to +\infty}{\sqrt{n}\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\sqrt{\frac{n}{2n+1}}\sqrt{2n+1}W_{2n+1}}=\frac{1}{\sqrt{2}}\times\sqrt{\frac{\pi}{2}}=\frac{\sqrt{\pi}}{2} $$
A: Redoing what has been done
many many many times before.
$\begin{array}\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
I_0
&=\int_0^1  dx\\
&= 1\\
I_1
&=\int_0^1 (1-x^2) dx\\
&=1-\dfrac13\\
&=\dfrac23\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
&=x(1-x^2)^n|_0^1+\int_0^1 2x^2n(1-x^2)^{n-1} dx\\
&\qquad\text{integrating by parts}\\
&\qquad f = (1-x^2)^n, f' = -2xn(1-x^2)^{n-1}, g' = 1, g = x\\
&=2n\int_0^1 x^2(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (x^2-1+1)(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-(1-x^2))(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-x^2)^{n-1} dx-2n\int_0^1 (1-x^2)^{n} dx\\
&=2nI_{n-1}-2nI_n\\
\text{so}\\
I_n
&=\dfrac{2n}{2n+1}I_{n-1}\\
\dfrac{I_n}{I_{n-1}}
&=\dfrac{2n}{2n+1}\\
I_n
&=\dfrac{I_n}{I_{0}}\\
&=\prod_{k=1}^n\dfrac{I_k}{I_{k-1}}\\
&=\prod_{k=1}^n\dfrac{2k}{2k+1}\\
&=\dfrac{\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k+1)}\\
&=\dfrac{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k+1)}\\
&=\dfrac{4^nn!^2}{(2n+1)!}\\
&=\dfrac{4^nn!^2}{(2n)!(2n+1)}\\
&\approx\dfrac{4^n(\sqrt{2\pi n}(n/e)^n)^2}{\sqrt{2\pi 2n}(2n/e)^{2n}(2n+1)}
\qquad\text{Stirling strikes twice}\\
&=\dfrac{4^n((2\pi n)(n^{2n}/e^{2n})}{2\sqrt{\pi n}4^nn^{2n}e^{2n}(2n+1)}\\
&=\dfrac{2\pi n}{2\sqrt{\pi n}(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{2n(1+1/(2n))}\\
&=\dfrac{\sqrt{\pi n}}{2n}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}(1-\dfrac1{2n}+O(\dfrac1{n^2}))\\
\end{array}
$
so
$\sqrt{n}I_n
=\dfrac{\sqrt{\pi }}{2}(1-\dfrac1{2n}+O(\dfrac1{n^2}))
\to\dfrac{\sqrt{\pi }}{2}
$.
