Estimating the integral $\sqrt{n}\cdot \int\limits_0^\pi \left( \frac{1 + \cos t}{2} \right)^n dt$ Consider the sequence $\{a_n\}$ defined by 
$$ a_n = \sqrt{n}\cdot \int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n dt.$$
An exercise in Rudin, Real and Complex Analysis, requires showing that this sequence is
convergent to a real number $a$, with $ a > 0$.
I don't have any idea of how to prove this. 
I only obtained the following estimation
$$ \begin{align*} 
\int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n  dt &= 2 \int_{0}^{\frac{\pi}{2}} \left( 1 - \sin^2 t \right)^n dt \\
 &> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - t^2)^n dt \\ 
&> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - n t^2) dt \\ &
= \frac{4}{3 \sqrt{n}}, 
\end{align*}$$
which shows that $ a_n > \frac{4}{3}$.
Thank you very much in advance for any help.
 A: $$
\begin{align}
\lim_{n\to\infty}\sqrt{n}\int_0^\pi\left(\frac{1+\cos(t)}{2}\right)^n\,\mathrm{d}t
&=\lim_{n\to\infty}\sqrt{n}\int_0^\pi\cos^{2n}(t/2)\,\mathrm{d}t\\
&=\lim_{n\to\infty}2\sqrt{n}\int_0^{\pi/2}\cos^{2n}(t)\,\mathrm{d}t\\
&=\lim_{n\to\infty}2\sqrt{n}\int_0^{n^{-1/3}}\left(1-\sin^2(t)\right)^n\,\mathrm{d}t\\
&=\lim_{n\to\infty}2\sqrt{n}\int_0^{n^{-1/3}}\left(1-t^2+\color{#C00000}{O}\big(t^4\big)\right)^n\,\mathrm{d}t\\
&=\lim_{n\to\infty}2\int_0^{n^{1/6}}\left(1-t^2/n+\color{#C00000}{O}\big(t^4/n^2\big)\right)^n\,\mathrm{d}t\\
&=\lim_{n\to\infty}2\int_0^{n^{1/6}}e^{-t^2+\color{#00A000}{O}(t^4/n)}\,\mathrm{d}t\\
&=2\int_0^\infty e^{-t^2}\,\mathrm{d}t\\[6pt]
&=\sqrt\pi
\end{align}
$$
Where, for $n\ge3$, the red $O$ has constant $1/3$ and the green $O$ has constant $2$.
Note that
$$
\begin{align}
\lim_{n\to\infty}\sqrt{n}\int_{n^{-1/3}}^{\pi/2}\left(1-\sin^2(t)\right)^n\,\mathrm{d}t
&\le\lim_{n\to\infty}\frac\pi2\sqrt{n}\left(1-\frac4{\pi^2}n^{-2/3}\right)^n\\
&\le\lim_{n\to\infty}\frac\pi2\sqrt{n}\,e^{-4n^{1/3}/\pi^2}\\[4pt]
&=0
\end{align}
$$

Motivation
As Didier points out, this approach is less elegant than those that make use of certain trigonometric identities, and perhaps Rudin posed this problem using trigonometric functions because one can use trigonometric identities to get a more elegant solution. However, I had computed the integral of $\cos^n(t)$ too many times recently, and I wanted to single out key sufficient features of $f$ that allow the limit
$$
\lim_{n\to\infty}\sqrt{n}\int_0^\pi f(t)^n\,\mathrm{d}t=\sqrt\pi
$$
those being


*

*$f(t)=1-t^2+O(t^4)$

*$f(t)\le1-kt^2$ for some $k\gt0$
A: Related problem. One can evaluate the integral using the beta function

$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2u-1}(\cos\theta)^{2v-1}\,d\theta, \qquad \mathrm{Re}(u)>0,\ \mathrm{Re}(v)>0. $$ 

First, we use the identity $1+\cos(x)=2\cos^2(x/2)$
$$ a_n=n^{1/2}\int_{0}^{\pi}\left(\frac{1+\cos(x)}{2}\right)^n dx = n^{1/2}\int_{0}^{\pi}{\cos^{2n}(x/2)} dx. $$
Then using the substitution $y=\frac{x}{2}$, we have
$$ a_n= 2 n^{1/2}\int_{0}^{\pi/2}{\cos^{n/2}(y)} dy=  \sqrt {\pi }{\frac {\sqrt {n}\,\Gamma  \left( n+1/2 \right)}{\Gamma 
 \left( n+1 \right) }}.$$
Taking the limit as $n\to \infty$, we get
$$ \lim_{n\to \infty} a_n = \sqrt{\pi}. $$
Note that, you can use Stirling approximation for $n!=\Gamma(n+1)\sim \left(\frac{n}{e}\right)^n\sqrt{ \pi n}. $ to evaluate the limit. 
A: Note that $a_n=\int\limits_0^\infty u_n(t)\mathrm dt$, where
$$
u_n(t)=2^{-n}(1+\cos(t/\sqrt{n}))^n\,\mathbf 1_{0\leqslant t\leqslant\pi\sqrt{n}}.
$$
It happens that $u_n\to u$ pointwise (can you show this?), where
$$
u(t)=\mathrm e^{-t^2/4},
$$
hence a natural conjecture is that $a_n\to a$, where
$$
a=\int_0^\infty u(t)\mathrm dt=\sqrt\pi.
$$
A tool to make sure this convergence happens is Lebesgue dominated convergence theorem, which requires to find some integrable $v$ such that $|u_n|\leqslant v$ for every $n$. It happens that $v=u$ fits the bill. 
To see why, note that $\cos(2t)+1=2\cos^2(t)$ hence $|u_n|\leqslant u$ for every $n$ as soon as, for every $x$ in $(0,\pi/2)$, $\cos(x)\leqslant\mathrm e^{-x^2/2}$. Any idea to show this last (classical) inequality?
A: Edited. The original answer is appended.
Put
$$c_m:=\int_0^{\pi/2} \cos^m s\ ds\qquad(m\geq0)\ .$$
Then $c_1=1$, $\> c_2={\pi\over4}$, and an easy partial integration shows that
$$c_m={m-1\over m} c_{m-2}\qquad(m\geq2)\ .\tag{1}$$ Now
$$a_n:=\sqrt{\mathstrut n}\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\sqrt{\mathstrut n}\int_0^{\pi/2}\cos^{2n}s\ ds$$
and therefore
$$a^2_n=4 n\> c_{2n}^2\ .$$
I claim that
$$a_n^2={c_{2n}\over c_{2n-1}}\pi\qquad(n\geq1)\ .\tag{2}$$
Proof by induction: $(2)$ is true for $n=1$. Assume that it is true for $n-1$. Using $(1)$ we then have
$$\eqalign{a_n^2 &= {n\over n-1}\left({c_{2n}\over c_{2n-2}}\right)^2 a_{n-1}^2 =
{n\over n-1}{2n-1\over 2n}{c_{2n}\over c_{2n-2}}{c_{2n-2}\over c_{2n-3}}\pi\cr &={n\over n-1}{2n-1\over 2n}{2n-2\over 2n-1}{c_{2n}\over c_{2n-1}}\pi={c_{2n}\over c_{2n-1}}\pi\ .\qquad\square\cr}$$
Since $c_m\leq c_{m-1}\leq c_{m-2}$ it follows easily from $(1)$ that $$\lim_{m\to\infty}{c_m\over c_{m-1}}=1\ .$$
From $(2)$ we therefore conclude that $\lim_{n\to\infty} a_n=\sqrt{\mathstrut \pi}$.
Original answer:
Let
$$b_n:=\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\int_0^{\pi/2}\cos^{2n}s\ ds\ .$$
Then
$$b_n =2\int_0^{\pi/2}\cos^{2n-2}s(1-\sin^2 s)\ ds=b_{n-1}-2\int_0^{\pi/2}\bigl(\cos^{2n-1}s\>\sin s\bigr)\cdot\sin s\ ds\ .$$
We integrate the last integral by parts and obtain
$$b_n=b_{n-1}-{1\over 2n-1} b_n\ .$$
Therefore the $b_n$ satisfy the recursion
$$b_n={2n-1\over 2n}\>b_{n-1}\qquad(n\geq1)\ ,$$
and the $a_n=\sqrt{\mathstrut n}\>b_n$ satisfy the recursion
$$a_n^2={(2n-1)^2\over 2n\>(2n-2)}\>a^2_{n-1}\qquad(n\geq2)\ .$$
As $a_1={\pi\over2}$ we therefore have
$$a_n^2=2\cdot\left({1\over2}\cdot{3\over2}\cdot{3\over4}\cdot{5\over4}\cdot{5\over6}\cdots{2n-1\over2n-2}{2n-1\over2n}\right)\cdot{\pi^2\over4}\qquad(n\geq2)\ .$$
In the large parenthesis the reciprocal of Wallis' product appears. It follows that this parenthesis  converges to ${2\over\pi}$, and we obtain
$$\lim_{n\to\infty}a_n=\sqrt{\mathstrut \pi}\ .$$
A: (too long to be a comment)
The integrals can be calculated exactly:
$$a_n=\frac{n^{1/2}}{2^n} \int_0^\pi(1+\cos t)^n dt. $$
For starters, we can use Newton's binomial: 
$$a_n=\frac{n^{1/2}}{2^n}\int_0^\pi \sum_{k=0}^n \binom{n}{k}\cos^ktdt. $$
Use the even-ness of the cosine around $t=\pi$:
$$a_n=\frac{n^{1/2}}{2^{n+1}}\sum_{k=0}^n \binom{n}{k} \int_0^{2\pi} \cos^k tdt $$
The integrals $\int_0^{2\pi} \cos^k t dt$ can be calculated using Euler's formula
$$\int_0^{2\pi} \cos^k t dt=\begin{cases} 0 & k=2m+1 \\ \frac{2\pi}{2^{2m}} \binom{2m}{m} & k=2m \end{cases}. $$
Therefore the summation is only needed for even indices $k=2m$, and
$$a_n=\frac{n^{1/2}}{2^{n+1}} \sum_{m=0}^{\lfloor{\frac{n}{2}} \rfloor} \binom{n}{2m} \frac{2\pi}{2^{2m}} \binom{2m}{m} $$
