# Definite Integral from 0 to 1 of x^4 (1 - x)^4 over (1 + x^2)

## Theorem

 $\ds \int_0^1 \dfrac {x^4 \paren {1 - x}^4} {1 + x^2} \rd x$ $=$ $\ds \dfrac {22} 7 - \pi$ $\ds$ $\approx$ $\ds 0 \cdotp 001264489$

## Proof

 $\ds \int_0^1 \dfrac {x^4 \paren {1 - x}^4} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \dfrac {x^4 \paren {1 - 4 x + 6 x^2 - 4 x^3 + x^4} } {1 + x^2} \rd x$ Fourth Power of Difference $\ds$ $=$ $\ds \int_0^1 \dfrac {x^4} {1 + x^2} \rd x - 4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x + 6 \int_0^1 \dfrac {x^6} {1 + x^2} \rd x - 4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x + \int_0^1 \dfrac {x^8} {1 + x^2} \rd x$ Linear Combination of Definite Integrals

We then establish a reduction formula:

 $\ds \dfrac {x^m} {1 + x^2}$ $=$ $\ds \dfrac {x^{m - 2} \paren {1 + x^2} - x^{m - 2} } {1 + x^2}$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds x^{m - 2} - \dfrac {x^{m - 2} } {1 + x^2}$

Then it is a matter of evaluating the individual integrals.

 $\ds \int_0^1 \dfrac {x^4} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \paren {x^2 - \dfrac {x^2} {1 + x^2} } \rd x$ from $(1)$ $\ds$ $=$ $\ds \int_0^1 x^2 \rd x - \int_0^1 \dfrac {x^2} {1 + x^2} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \intlimits {\dfrac {x^3} 3} 0 1 - \int_0^1 \dfrac {x^2} {1 + x^2} \rd x$ Primitive of Power $\ds$ $=$ $\ds \intlimits {\dfrac {x^3} 3} 0 1 - \bigintlimits {x - \arctan x} 0 1$ Primitive of $\dfrac {x^2} {x^2 + a^2}$ $\ds$ $=$ $\ds \dfrac 1 3 - \paren {1 - \arctan 1} - \paren {0 - \arctan 0}$ plugging in the limits $\text {(2)}: \quad$ $\ds$ $=$ $\ds \dfrac \pi 4 - \dfrac 2 3$ simplifying

 $\ds \int_0^1 \dfrac {x^5} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \paren {x^3 - \dfrac {x^3} {1 + x^2} } \rd x$ from $(1)$ $\ds$ $=$ $\ds \int_0^1 x^3 \rd x - \int_0^1 \dfrac {x^3} {1 + x^2} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \intlimits {\dfrac {x^4} 4} 0 1 - \int_0^1 \dfrac {x^3} {1 + x^2} \rd x$ Primitive of Power $\ds$ $=$ $\ds \intlimits {\dfrac {x^4} 4} 0 1 - \intlimits {\frac {x^2} 2 - \frac {a^2} 2 \, \map \ln {x^2 + a^2} } 0 1$ Primitive of $\dfrac {x^3} {x^2 + a^2}$ $\ds$ $=$ $\ds \dfrac 1 4 - \paren {\paren {\dfrac 1 2 - 0} - \paren {\frac 1 2 \, \map \ln {1 + 1^2} - \frac 1 2 \, \map \ln {1 + 0^2} } }$ plugging in the limits $\ds$ $=$ $\ds \dfrac 1 4 - \dfrac 1 2 + \frac 1 2 \ln 2$ simplifying $\text {(3)}: \quad$ $\ds$ $=$ $\ds \dfrac {\ln 2} 2 - \dfrac 1 4$ simplifying $\ds \leadsto \ \$ $\ds -4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x$ $=$ $\ds 1 - 2 \ln 2$

 $\ds \int_0^1 \dfrac {x^6} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \paren {x^4 - \dfrac {x^4} {1 + x^2} } \rd x$ from $(1)$ $\ds$ $=$ $\ds \int_0^1 x^4 \rd x - \int_0^1 \dfrac {x^4} {1 + x^2} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \intlimits {\dfrac {x^5} 5} 0 1 - \int_0^1 \dfrac {x^4} {1 + x^2} \rd x$ Primitive of Power $\ds$ $=$ $\ds \intlimits {\dfrac {x^5} 5} 0 1 - \paren {\dfrac \pi 4 - \dfrac 2 3}$ from $(2)$ $\ds$ $=$ $\ds \dfrac 1 5 - \dfrac \pi 4 + \dfrac 2 3$ plugging in the limits $\text {(4)}: \quad$ $\ds$ $=$ $\ds \dfrac {13} {15} - \dfrac \pi 4$ simplifying $\ds \leadsto \ \$ $\ds 6 \int_0^1 \dfrac {x^6} {1 + x^2} \rd x$ $=$ $\ds \dfrac {26} 5 - \dfrac {3 \pi} 2$

 $\ds \int_0^1 \dfrac {x^7} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \paren {x^5 - \dfrac {x^5} {1 + x^2} } \rd x$ from $(1)$ $\ds$ $=$ $\ds \int_0^1 x^5 \rd x - \int_0^1 \dfrac {x^5} {1 + x^2} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \intlimits {\dfrac {x^6} 6} 0 1 - \int_0^1 \dfrac {x^5} {1 + x^2} \rd x$ Primitive of Power $\ds$ $=$ $\ds \intlimits {\dfrac {x^6} 6} 0 1 - \paren {\dfrac {\ln 2} 2 - \dfrac 1 4}$ from $(3)$ $\ds$ $=$ $\ds \dfrac 1 6 - \dfrac {\ln 2} 2 + \dfrac 1 4$ plugging in the limits $\ds$ $=$ $\ds \dfrac 5 {12} - \dfrac {\ln 2} 2$ simplifying $\ds \leadsto \ \$ $\ds -4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x$ $=$ $\ds 2 \ln 2 - \dfrac 5 3$

 $\ds \int_0^1 \dfrac {x^8} {1 + x^2} \rd x$ $=$ $\ds \int_0^1 \paren {x^6 - \dfrac {x^6} {1 + x^2} } \rd x$ from $(1)$ $\ds$ $=$ $\ds \int_0^1 x^6 \rd x - \int_0^1 \dfrac {x^6} {1 + x^2} \rd x$ Linear Combination of Definite Integrals $\ds$ $=$ $\ds \intlimits {\dfrac {x^7} 7} 0 1 - \int_0^1 \dfrac {x^6} {1 + x^2} \rd x$ Primitive of Power $\ds$ $=$ $\ds \intlimits {\dfrac {x^7} 7} 0 1 - \paren {\dfrac {13} {15} - \dfrac \pi 4}$ from $(4)$ $\ds$ $=$ $\ds \dfrac 1 7 - \dfrac {13} {15} + \dfrac \pi 4$ plugging in the limits

It remains to gather up the terms.

First the terms in $x^5$ and $x^7$, as there is some important cancelling out:

 $\ds \paren {-4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x} + \paren {-4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x}$ $=$ $\ds \paren {1 - 2 \ln 2} + \paren {2 \ln 2 - \dfrac 5 3}$ $\ds$ $=$ $\ds 1 - \dfrac 5 3$ $\ds$ $=$ $\ds -\dfrac 2 3$

Thus we continue:

 $\ds \int_0^1 \dfrac {x^4 \paren {1 - x}^4} {1 + x^2} \rd x$ $=$ $\ds \paren {\dfrac \pi 4 - \dfrac 2 3} - \dfrac 2 3 + \paren {\dfrac {26} 5 - \dfrac {3 \pi} 2} + \paren {\dfrac 1 7 - \dfrac {13} {15} + \dfrac \pi 4}$ $\ds$ $=$ $\ds \paren {-\dfrac 2 3 - \dfrac 2 3} + \dfrac 1 7 + \dfrac {26} 5 - \dfrac {13} {15} - \pi$ rearranging, and resolving the multiples of $\pi$ $\ds$ $=$ $\ds \dfrac 1 7 + \dfrac {78 - 13 - 10 - 10} {15} - \pi$ gathering all the multiples of $\dfrac 1 {15}$ $\ds$ $=$ $\ds 3 + \dfrac 1 7 - \pi$ arithmetic

The result follows.

$\blacksquare$