Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
 A: Another solution is given by the integral
$$
0 < \int_0^1 \frac{x^4(1-x)^8}{4(1+x^2)}dx = \pi -\frac{2419}{770} = \pi - \frac{333}{106}-\frac{1}{20405}
$$
This proves the stricter condition
$$
\pi > \frac{333}{106}+\frac{1}{20405}
$$
which implies
$$
\pi > \frac{333}{106}
$$
Similarly, for the fourth convergent (formula (6) http://www.math.ucla.edu/~vsv/resource/general/Lucas.pdf)
$$0<\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx=\frac{3849155}{1225224}-\pi=\frac{355}{113}-\frac{5}{138450312}-\pi$$
Therefore
$$\pi<\frac{355}{113}-\frac{5}{138450312}$$
and
$$\pi<\frac{355}{113}$$
Even for the first convergent
$$0<2\int_0^1 \frac{x(1-x)^2}{(1+x^2)}dx=\pi-3$$
so
$$\pi>3$$
(See https://math.stackexchange.com/a/1618454/134791 for a proof for $3<\pi<4$ that uses this integral)
A series proof that $\pi>\frac{333}{106}$ is given by

$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106}$$

A: Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the Wikipedia page about this integral. For instance, $$0 < \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$
In general, 
$$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx
<\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx
<\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$
which for $n=1$ (the integral in the question) gives slightly better bounds than just $\pi < 22/7$:
$$ \frac{1}{1260} < \frac{22}{7} - \pi < \frac{1}{630}$$
A: 
This integral would do the job:

$$\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{530(1+x^2)}\:dx= \pi -\frac{333}{106}$$


*

*Also you can refer to S.K. Lucas Integral proofs that $355/113 > \pi$, Gazette, Aust. Math. Soc.  32 (2005), 263-266.


*This is the link. (Thanks to lhf for pointing out.)

A: From the relationship 
$$\frac{333}{106}=\frac{377-2·22}{120-2·7}$$
and integrals
$$\pi = \frac{22}{7} - \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
and
$$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$
we obtain
$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(1-x)^2)}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = 106\pi-333$$
Therefore,

$$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$

The numerator can be shown to be nonnegative for $0\leq x\leq 1$.
In fact, we can get a smaller numerator that is still nonnegative, which leads to a closer approximation to $\pi$

$$\pi=\frac{21991}{7000} +\frac{1}{50} \int_0^1 \frac{x^4 (1 - x)^4 \left(4 - 27 x (1 - x)^2\right)}{\left(1 + x^2\right)} dx$$

Here the numerator has been adjusted to have a double zero in $(0,1)$ while not crossing the axis WA link
The resulting fraction is
$$\frac{21991}{7000} = \frac{22}{7}-\frac{9}{7000}$$
which allows for writing the double inequality
$$\frac{22}{7}-\frac{9}{7000}<\pi<\frac{22}{7}$$
or, equivalently, an upper bound for the error in Archimedes' approximation
$$\frac{22}{7}-\pi<\frac{9}{7000}$$
A: Let us consider the polynomial
$$P_n(x):=1-x^2+x^4-x^6+\cdots x^{2n-2}=\frac{x^{2n}+1}{x^2+1}.$$
We have
$$0<\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx<\int_0^1\left(\frac{x^{2n}}{0+1}\right)^2dx=\frac1{4n+1},$$ and the integral can be made as small as desired.
On another hand, the remainder of the long division of $(P_n(x)-1)^2$ by $(x^2+1)^2$ is a binomial $ax^2+b$, with $a,b$ integer. Then
$$\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx=\int_0^1\left(Q(x)+\frac{ax^2+b}{(x^2+1)^2}\right)dx.$$
As
$$\int_0^1\frac{ax^2+b}{(x^2+1)^2}dx=\frac{b-a}4+\frac{b+a}8\pi$$
we can get arbitrarily close rational approximations by a rational integral.

For example, with $P_2:=1-x^2+x^4$ we have
$$\frac{(x^4-x^2)^2}{(x^2+1)^2}=x^4-4x^2+8-\frac{12x^2+8}{(x^2+1)^2}$$ so that by integration
$$0<\frac15-\frac43+8-\frac{5\pi}2+1<\frac19$$ or
$$\frac{698}{225}<\pi<\frac{236}{75}.$$
A: Without restrictions on the integrals, the question is virtually meaningless, because you can take any known bounds on $\pi$ (such as terms of the Gregory series), say $a<\pi<b$ and write
$$\int_0^a dx<\pi<\int_0^bdx.$$
Or if the function must be non-trivial, use any convergent squeezing of $\dfrac4{x^2+1}$ and
$$\int_0^1f(x)\,dx<\int_0^1\frac{4\,dx}{x^2+1}=\pi<\int_0^1g(x)\,dx$$ that you can make as tight as wanted.
Or better, any convergent squeezing of $\pi$, and
$$\int_0^1f(x)\,dx<\pi<\int_0^1g(x)\,dx.$$
In my other answer, I showed constructively that that there are bracketings of $\pi$ with integrals of rational functions as tight as you want, even though they are pretty inefficient.
