Find limit $a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$

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 $\pi$.