Prove that the following sequence converges Prove that for every $a_0\in(0,2\pi)$ the following sequence
$$
a_{n+1} = \int_0^{a_n}(1+\frac{1}{4}\cos^{2n+1}t) \,\mathrm{d}t
$$
converges and find a limit of this sequence.
It is evident that this sequence is bounded. But it isn't monotonic. Also I have tried to show that this sequence is fundamental, but my attempts failed.
 A: This is only a partial answer - it shows that the sequence converges, but does not give the limit.
Define
$$I_n(x) = \int_0^x \cos^{2n+1}(t)dt .$$
We have $I_n(\pi) = 0$ because $\cos^{2n+1}(\pi/2 + t) = -\cos^{2n+1}(\pi/2 - t)$. For $a \in (0,2\pi)$ define $$f(n,a) = \int_0^a (1 + 1/4\cos^{2n+1}(t))dt = a + 1/4 I_n(a) .$$
Then $a_{n+1} = f(n,a_n)$. For $a_0 = \pi$ we get $a_n = \pi$ for all $n$; this sequence trivially converges to $\pi$. We claim that
$$a < f(n,a) <\pi \text{ for } 0 < a < \pi .$$
This implies that $(a_n)$ is bounded and strictly increasing, i.e. is convergent. The case $\pi < a < 2\pi$ can be treated similarly (we get $\pi < f(n,a) < a$ so that $(a_n)$ is bounded and strictly decreasing).
Let us prove the above claim. For $0 < a \le \pi/2$ we have $0 < I_n(a) < a \le \pi/2$ which holds because $0 \le \cos^{2n+1}(t) \le 1$ for $0 \le t \le \pi/2$. For $\pi/2 < a < \pi$ we have $I_n(a) = I_n(\pi) - \int_a^\pi \cos^{2n+1}(t)dt  = - \int_a^\pi \cos^{2n+1}(t)dt = \int_a^\pi \lvert \cos^{2n+1}(t) \rvert dt \in (0, \pi - a)$.
Added: For $a < b$ we have $f(n,a) < f(n,b)$ because $f(n,b) - f(n,a) = b - a + 1/4\int_a^b \cos^{2n+1}(t)dt \ge b -a -1/4\int_a^b \lvert \cos^{2n+1}(t) \rvert dt >$ $ b -a - 1/4(b-a) > 0$.
Letting denote $\overline{a}_0$ the limit of the sequence $(a_n)$ starting with $a_0$ we see that $\overline{a}_0 \le \overline{b}_0$ when $a_0 < b_0$.
