Find limit $a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$ Find the limit of the sequence:
$$a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$$
such that $a_0 \in (0, 2 \pi)$

That was one of the tasks in the Olympiad.
Here is my approach.
First, I wanted to simplify the integral:  
$\int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt = \int_{0}^{a_n}(dt) +  \frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt$
That leads to the following relation:
$$a_{n + 1} = a_n + \frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt$$
Now, there is a $\cos t$ with some power which reminded me of the standart integral $\int \cos^n(x) dx$. We can find a recursive formula for it in the following way:
$I_n = \int \cos^n(x) dx = \int \cos(x) \cos^{n - 1}(x) dx = \sin x \cos^{n - 1}x - (n - 1)\int \sin(x) \cos^{n - 2}(x) (- \sin (x)) dx.$
This leads to
$I_n = \sin x \cos^{n - 1}x + (n - 1) I_{n - 2} - (n - 1) I_n$  
And final recurrence relation is
$$I_n = \frac{1}{n} \sin x \cos^{n - 1}x + \frac{n - 1}{n} I_{n - 2}$$
For a long time I am trying to make a connection between the original integral $\int_{0}^{a_n} \cos^{2n + 1} (t) dt$ and this recurrence relation, but I have failed to come up with anything meaningful at the moment.

Well, I guess we can just plug in $2n + 1$ instead of $n$ and we get
$$I_{2n + 1} = \frac{1}{2n + 1} \sin x \cos^{2n}x + \frac{2n}{2n + 1} I_{2n - 1}$$

Ok, now if we try to evaluate this as definite integral we should get  
$I_{2n + 1}(a_n) - I_{2n + 1}(0) = (\frac{1}{2n + 1} \sin a_n \cos^{2n}a_n + \frac{2n}{2n + 1} I_{2n - 1}(a_n)) - (0 + \frac{2n}{2n + 1} I_{2n - 1}(0))$
$I_{2n + 1}(a_n) - I_{2n + 1}(0) = \frac{1}{2n + 1} \sin a_n \cos^{2n}a_n + \frac{2n}{2n + 1} I_{2n - 1}(a_n) - \frac{2n}{2n + 1} I_{2n - 1}(0).$
So, 
$$\frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt = \frac{1}{4(2n + 1)} \sin a_n \cos^{2n}a_n + \frac{2n}{4(2n + 1)} \big[ I_{2n - 1}(a_n) - I_{2n - 1}(0) \big] $$
$$\frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt = \frac{1}{4(2n + 1)} \sin a_n \cos^{2n}a_n + \frac{2n}{4(2n + 1)} \big[ I_{2n - 1}(a_n) - \cos a_0 \big] $$
I would appreciate any help if you provide me with some insights or clues on how to proceed.
 A: The limit is $\pi$.
First note that
$$a_{n+1} = a_n + \frac{1}{4}\int_0^{a_n} \cos^{2n+1} t\,dt.$$
$\cos^{2n+1} t$ is positive on $(0,\pi/2)$, negative on $(\pi/2,3\pi/2)$, and again positive on $(3\pi/2,2\pi)$. By the symmetries of the cosine, we have
$$\int_0^{\pi} \cos^{2n+1} t\,dt = 0 = \int_0^{2\pi} \cos^{2n+1} t\,dt,$$
so $a_{n+1} = a_n$ if $a_n = \pi$, $a_{n+1} > a_n$ if $0 < a_n < \pi$, and $a_{n+1} < a_n$ if $\pi < a_n < 2\pi$. Further, for $\pi < a_n < 2\pi$ we have
$$a_{n+1} - a_n = \frac{1}{4}\int_{\pi}^{a_n} \cos^{2n+1} t\,dt > \frac{1}{4} \int_{\pi}^{a_n} (-1)^{2n+1}\,dt = -\frac{1}{4}(a_n-\pi),$$
and similarly for $0 < a_n < \pi$ we have
$$a_{n+1} - a_n = \frac{1}{4}\int_0^{a_n} \cos^{2n+1} t\,dt = -\frac{1}{4}\int_{a_n}^{\pi} \cos^{2n+1} t\,dt < -\frac{1}{4} \int_{a_n}^{\pi} (-1)^{2n+1}\,dt = \frac{1}{4}(\pi - a_n),$$
so $a_n > \pi \implies a_{n+1} > \pi$, and $a_n < \pi \implies a_{n+1} < \pi$.
Thus, whatever the starting point, the sequence is monotonic and bounded, hence it converges.
Consider the case $0 < a_0 < \pi$, the case $\pi < a_0 < 2\pi$ is analogous.
For the sake of contradiction, suppose that $a = \lim\limits_{n\to\infty} a_n < \pi$.
If $a \leqslant \pi/2$, then for all large enough $n$ we would have
\begin{align}
a_{n+1} - a_{n} &> \frac{1}{4} \int_0^{\frac{1}{2n+1}} \cos^{2n+1} t\,dt \\
&> \frac{1}{4}\int_0^{\frac{1}{2n+1}} (1-t^2)^{2n+1}\,dt \\
&> \frac{1}{4} \int_0^{\frac{1}{2n+1}} 1 - (2n+1)t^2\,dt \\
&= \frac{1}{4}\biggl(\frac{1}{2n+1} - \frac{1}{3(2n+1)^2}\biggr) \\
&> \frac{1}{8(2n+1)},
\end{align}
but the series
$$\sum_{n = 1}^{\infty} \frac{1}{8(2n+1)}$$
is divergent, while $\sum (a_{n+1} - a_n)$ is convergent. Thus we have a contradiction.
If we assume $\pi/2 < a < \pi$, we get the analogous contradiction by noting that for all large enough $n$ we'd have
$$a_{n+1} - a_n > -\frac{1}{4}\int_{\pi - \frac{1}{2n+1}}^{\pi} \cos^{2n+1} t\,dt > \frac{1}{8(2n+1)}.$$
Thus the limit is $\pi$.
A: You can easily verify that $$I_(2n+1) \to 0$$,$$ as n\to \infty$$
And $$\lim_{n\to\infty} a_(n+1)-a_n =0$$
And a_1=\int_{0}{a_0} (1+cost)dt=2π
