The value of $\int_0^{2\pi}\cos^{2n}(x)$ and its limit as $n\to\infty$ 
Calculate $I_{n}=\int\limits_{0}^{2\pi} \cos^{2n}(x)\,{\rm d}x$
  and show that  $\lim_{n\rightarrow \infty} I_{n}=0$

Should I separate $\cos^{2n}$ or I should try express it in Fourier series?

Here is a completely 
elementary 
(i.e., nothing beyond basic integration)
proof.
Taking advantage of
the symmetries of
$\cos$,
$I_{n}
=\int\limits_{0}^{2\pi} \cos^{2n}(x)\,{\rm d}x
=4\int\limits_{0}^{\pi/2} \cos^{2n}(x)\,{\rm d}x
=4\int\limits_{0}^{\pi/2} (\cos^2(x))^{n}\,{\rm d}x
=4\int\limits_{0}^{\pi/2} (1-\sin^2(x))^{n}\,{\rm d}x
$.
Since
$\sin(x)
\ge 2 x/\pi
$
on
$0 \le x \le \pi/2$,
$I_{n}
\le 4\int\limits_{0}^{\pi/2} (1-(2x/\pi)^2)^{n}\,{\rm d}x
= 2\pi\int\limits_{0}^{1} (1-x^2)^{n}\,{\rm d}x
= 2\pi\int\limits_{0}^{1} (1-x^2)^{n}\,{\rm d}x
$.
Split the integral into two part,
$\int_0^d$ and $\int_d^1$.
In the first part,
since the integrand
is at most $1$,
the integral
is at most $d$.
In the second part,
the integrand 
is at most
$(1-d^2)^n$,
so the integral is less than
$(1-d^2)^n$.
We now want to relate $d$ and $n$
so both integrals are small.
To make
$(1-d^2)^n
< c
$,
where
$0 < c < 1$,
we want
$n\ln(1-d^2)
< \ln c
$
or
$n(-\ln(1-d^2))
> (-\ln c)
$
or
$n
> \frac{-\ln c}{-\ln(1-d^2)}
$.
Therefore,
for any positive
$c$ and $d$,
by choosing
$n
> \frac{-\ln c}{-\ln(1-d^2)}
$
we can make
$I_n
<
2\pi(d+c)
$.
By choosing $c$ and $d$
arbitrarily small,
so is $I_n$,
so $\lim_{n \to \infty} I_n
= 0$.
To get a more elementary
bound on $n$,
since
$-\ln(1-z)
>z
$
if $0 < z < 1$,
$\frac{-\ln c}{-\ln(1-d^2)}
<\frac{-\ln c}{d^2}
$.
so choosing
$n > \frac{-\ln c}{d^2}$
will do.
To completely eliminate $\ln$s
in the bound for $n$
set $c = 10^{-m}$.
We get
$I_n < 2\pi(d+10^{-m})$
by choosing
$n
>\frac{m \ln 10}{d^2}
$.
