<s>[INST] Evaluate the integrals in $\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$ Note that when $0\le x \le 1$ we have $$\frac 12 \le \frac 1 {1+x^2} \le 1.$$
Hence, $$\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$$
Evaluate the integrals above to show that $$\frac1{1260}\le\frac {22}7-\pi\le\frac1{630}.$$
All the integrands include the expression $x^4(1-x)^4$ but I'm not sure how to continue.
 [/INST] To evaluate the first and third integrals, simply expand:
$$x^4 (1 - x)^4 = x^4 - 4 x^5 + 6 x^6 - 4 x^7 + x^8.$$ Then, integrate term by term.
To evaluate the second integral, apply polynomial long division, which will give that the middle integrand has the form
$$\frac{x^4 (1 - x)^4}{1 + x^2} = p(x) - \frac{4}{1 + x^2},$$
where $p(x)$ is some polynomial.
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