<s>[INST] 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.
 [/INST] 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
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