Evaluate the integrals in $\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$

Question

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.

Answer 1

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.

Answer 2

The first and third integrals can be solved using the substitution $x=\sin^2(\theta)$. The integral will take the form $2\displaystyle\int_0^{\pi/2}\sin^{2a-1}(\theta)\cos^{2b-1}(\theta)d\theta$, which is equal to the beta function $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. The middle integral can be split into fractions according to Travis's answer.