Problem source

Find constants $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$,  $a_{6}$  and $b$ such that
\[x^{4}(1-x)^{4}=(a_{6}x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+
a_{2}x^{2}+a_{1}x+a_{0})(x^{2}+1)+b.\]
Hence, or otherwise, prove that
\[\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}{\rm d}x
=\frac{22}{7}-\pi.\]
Evaluate $\displaystyle{\int_{0}^{1}x^{4}(1-x)^{4}{\rm d}x}$
and deduce that
\[\frac{22}{7}>\pi>\frac{22}{7}-\frac{1}{630}.\]

Solution source

Plugging in $x = i$ we obtain $(1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4$.

\begin{align*}
x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\
&= x^8-4x^7+6x^6-4x^5+x^4 \\
&= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\
&= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\
&= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\
&= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\
&= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\
&= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\
\end{align*}

So
\begin{align*}
\int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\
&= \frac17 - \frac46+1-\frac43+4 - \pi \\
&= \frac{22}7 - \pi
\end{align*}

\begin{align*}
\int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\
&= \frac{4!4!}{9!} \\
&= \frac1{630}
\end{align*}

Therefore since $0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4$ we must have that

\begin{align*}
&& 0 &< \frac{22}7 - \pi \\
\Rightarrow && \pi & < \frac{22}{7} \\
&& \frac{22}{7} - \pi &< \frac1{630} \\
 \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi
\end{align*}

which is what we wanted.