William Lowell Putnam Integral Problem Prove That
$$
\frac{22}{7}-\pi
=
\int_{0}^{1}\frac{x^{4}\left(1 - x\right)^{4}}{1 + x^{2}}\,{\rm d}x
$$
 A: Literally all you have to do is expand it out, divide, and integrate.
$$
x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4
$$
Now use long division:
$$
x^4(1-x)^4 = x^6(x^2+1)- 4x^5(x^2+1) + 5x^4(x^2 + 1) - 4x^2(x^2+1)+4(x^2+1) -4 \\
\frac{x^4(1-x)^4}{1+x^2} = (x^6 -4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{x^2+1} \\
$$
We integrate this:
$$
\int_0^1 \! (x^6 -4x^5 + 5x^4 - 4x^2 + 4) \,\mathrm{d}x - \int_0^1 \! \frac{4}{x^2+1} \, \mathrm{d}x \\
= \frac{22}{7} - 4(\arctan(1) - \arctan(0)) \\
= \frac{22}{7} - \pi
$$
A: I Tried in this way: 
$$ \frac{x^4\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1+1\right)\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1\right)\,\left(1-x\right)^4}{1+x^2}+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x^2-1\right)\left(1-x\right)^4+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x+1\right)\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x-1\right)^6+2\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}$$ First two terms are straight forward to integrate.
$$\int_0^1 \frac{\left(1-x\right)^4}{1+x^2}=\int_0^1\left(x^2-4x+5\right)-\frac{4}{x^2+1}=\frac{10}{3}-\pi$$
Can any one suggest ant alternate ways to do this problem..
A: $$\begin{align}\int_0^1 dx \frac{x^4 (1-x)^4}{1+x^2} &= \int_0^1 dx \frac{x^4-4 x^5 + 6 x^6 - 4 x^7 + x^8}{1+x^2}\\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{4+2 k}-4 x^{5+2 k}+6 x^{6+2 k}-4 x^{7+2 k}+x^{8+2 k}\right )\\&= \sum_{k=0}^{\infty} (-1)^k \left (\frac1{5+2 k}-\frac{4}{6+2 k}+\frac{6}{7+2 k}-\frac{4}{8+2 k}+\frac1{9+2 k} \right ) \end{align} $$
Now,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{5+2 k} = \frac15-\frac17+\frac19-\cdots=\frac{\pi}{4}-\left (1-\frac13 \right )  $$
Similarly,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{7+2 k} = \frac17-\frac19+\frac1{11}-\cdots=\left (1-\frac13+\frac15 \right ) -\frac{\pi}{4} $$
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{9+2 k} = \frac19-\frac1{11}+\frac1{13}-\cdots=\frac{\pi}{4}-\left (1-\frac13+\frac15-\frac17 \right )  $$
Meanwhile,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{3+ k} = \frac13-\frac14+\frac15-\cdots=\log{2}-\left (1-\frac12 \right )$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{4+ k} = \frac14-\frac15+\frac16-\cdots=\left (1-\frac12+\frac13 \right )-\log{2}$$
The result is, summing all of the above numbers,
$$-\left (1-\frac13 \right )+ 6 \left (1-\frac13+\frac15 \right )-\left (1-\frac13+\frac15-\frac17 \right )- \frac{2}{3} - \pi = \frac{22}{7}-\pi$$
A: $\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx\;,$$ Let $\displaystyle x= \tan \phi\;,$ Then $dx = \sec^2 \phi.$
And Changing Limit, We Get $$\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^4 \phi\cdot \left(1-\tan \phi\right)^4d\phi.$$
$$\displaystyle  =\int_{0}^{\frac{\pi}{4}}\tan^4 \phi \cdot \left(1-4\tan \phi+6\tan^2 \phi-4\tan^3 \phi+\tan^4 \phi\right).$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\tan^4 \phi-4\tan^5 \phi+6\tan^6 \phi-4\tan^7 \phi+\tan^8 \phi \right\}d\phi.$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\left[\tan^4 \phi+\tan^6 \phi \right]-4\left[\tan^5 \phi \cdot \sec^2 \phi \right]+\left[\tan^6 \phi+\tan^8 \phi  \right]+4\tan^6 \phi \right\}d\phi.................(\star)\color{\red}\checkmark.$$
Now Let $$\displaystyle J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n}\phi d\phi\;,$$ Then $$\displaystyle J_{n+2} = \int_{0}^{\frac{\pi}{4}}\tan^{n+2}\phi d\phi$$
Now $$\displaystyle J_{n+2}+J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n} \phi \cdot \sec^2 \phi d\phi  = \frac{1}{n+1}\Rightarrow J_{n+2}+J_{n} = \frac{1}{n+1}............(\star \star)\color{\red}\checkmark.$$
Now at $n=0\;,$ we get $\displaystyle J_{0} = \frac{\pi}{4}\;,$ Then $\displaystyle J_{2} = \left(1-\frac{\pi}{4}\right)$ and $\displaystyle $ When $ n= 2\;,$ We get $$\displaystyle J_{4} = \left(\frac{\pi}{4}-\frac{2}{3}\right).$$
and When $n=4\;,$ We get $\displaystyle J_{6} = \left(\frac{13}{15}-\frac{\pi}{4}\right).$ and We get $\displaystyle J_{4}+J_{6}=\frac{1}{5}$ And $\displaystyle J_{6}+J_{8} = \frac{1}{7}$ 
Now Put these values in $............(\star)\color{\red}\checkmark.\;,$ We get
$$\displaystyle I = \frac{1}{5}-\frac{2}{3}+\frac{1}{7}+\frac{52}{15}-\pi = \left(\frac{22}{7}-\pi\right).$$
