# Shortest method for $\displaystyle\int_{0}^{1}\dfrac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$

I don’t want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is $\pi$ so tried several trigonometric substitutions too

#### Solutions Collecting From Web of "Shortest method for $\displaystyle\int_{0}^{1}\dfrac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$"

It is not that difficult to expand using the binomial theorem, and some terms drop out nicely.

$$x^4(1-x)^4=x^4-4x^5+6x^6-4x^7+x^8$$

Now look at the odd powers $-4x^5-4x^7=-4x^5(1+x^2)$

And then you can use polynomial division with the even powers only – which simplifies things greatly.

Note that $(1 \pm i)^4 = -4$, hence $(1-x)^4 + 4 = p(x)(1+x^2)$ for some normalized second degree polynomial $p$. To find $p$, we plug in $0$, $1$, giving
$$5 = p(0), 4 = 2p(1)$$
With the ansatz $p(x) = x^2 + ax + b$, we have
$$5 = b,\quad 2+2a + 2b = 4 \iff a = -4$$
Hence
$$\frac{(1-x)^4}{1+x^2} = x^2 – 4x + 5 – \frac{4}{1+x^2}$$
Multiplying by $x^4$ gives
$$\frac{x^4(1-x)^4}{1+x^2} = (x^6 – 4x^5 + 5x^4) – \frac{4x^4}{1+x^2}$$
Now we use
$$\frac{x^4}{1+x^2} = \frac{x^4 – 1}{x^2 + 1} + \frac{1}{x^2+ 1} = x^2 – 1 + \frac{1}{x^2 + 1}$$
giving
$$\frac{x^4(1-x)^4}{1+x^2} = x^6 – 4x^5 + 5x^4 – 4x^2 + 4 – \frac 4{x^2 + 1}$$
So
$$\int \frac{x^4(1-x)^4}{1+x^2}\, dx = \frac{x^7}7 – \frac 23x^6 + x^5 – \frac 43x^3 + 4x – 4\arctan x.$$

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

$$\displaystyle I = \frac{1}{5}-\frac{2}{3}+\frac{1}{7}+\frac{52}{15}-\pi = \left(\frac{22}{7}-\pi\right).$$

$f(z)=\frac{z^4(1-z)^4}{1+z^2}$ is a meromorphic function with simple poles at $z=\pm i$. Since

$$\text{Res}_{z=\pm i}\frac{z^4(1-z)^4}{(z+i)(z-i)}=\lim_{z\to \pm i}\frac{z^4(1-z)^4}{z\pm i}=\pm 2i \tag{1}$$
we have that
$$f(z)-\left(\frac{2i}{z-i}+\frac{-2i}{z+i}\right) = f(z)+\frac{4}{z^2+1}\tag{2}$$
is an entire function: a polynomial, in particular. It follows that $z^4(1-z)^4+4$ is a multiple of $z^2+1$. Not surprising, since by Sophie Germain’s identity
$$w^4+4 = (w^2-2w+2)(w^2+2w+2) \tag{3}$$
hence:
$$z^4(1-z)^4+4 = (1+z^2)(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\tag{4}$$
as well as:
$$\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx = -\pi+\int_{0}^{1}(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\,dz.\tag{5}$$