Integrals

$\int_0^1 \! \frac{x^7}{\sqrt{1-x^2}} \, \mathrm{d}x.$

$u = \sqrt{ 1-x^2 } \\ u^2 = 1 - x^2 \\ 2udu = -2xdx$

$\int_0^1 \! \frac{x^6}{u} *x \, \mathrm{d}x.\\ \int_1^0 \! \frac{x^6}{u} *-u \, \mathrm{d}u.\\ \int_0^1 \! x^6 \, \mathrm{d}u.\\ \int_0^1 \! (1-u^2)^3 \, \mathrm{d}u.\\ \int_0^1 \! 1 - 3u^2 + 3u^4 - u^6 \, \mathrm{d}u.$

which is $1 - 1 + \frac{3}{5} - \frac{1}{7} = .457$

Relevant wiki: Integration Techniques

Let $I =\int _{ 0 }^{ 1 }{ \frac { { x }^{ 7 } }{ \sqrt { \left( 1-{ x }^{ 2 } \right) } } dx }$ just substitue $x=\sin { z }$ .Hence $dx=coszdz$ . Transforming the integral to:- $I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { \left( \sin { z } \right) }^{ 7 } }{ \cos { z } } } \cos { z } dz\Rightarrow I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \sin { z } \right) }^{ 7 }dz }$ Now we can simply use the reduction formula $\displaystyle { I }_{ n }=\frac { n-1 }{ n } { I }_{ n -2}\quad \left\{ { I }_{ n }=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \sin { x } \right) }^{ n }dx } \right\}$ . And get $I=\frac{16}{35} \approx 0.457$ . Or else you can also use the Beta function Like Sir Chew-Seong Cheong has used

$\begin{aligned} I & =\int_0^1 \frac {x ^7}{\sqrt {1 - x^2}} dx\\ & =\int_0^1 x ^7(1 - x^2) ^{-\frac 12}\ dx \\ & = \frac 12 B \left(4,\frac 12 \right) \\&= \frac {\Gamma(4) \Gamma \left( \frac 12 \right)} {2 \Gamma \left (\frac 92\right)} \\ & = \frac {3!\Gamma \left( \frac 12 \right)} { 2 \cdot \frac7 2 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12 \Gamma \left( \frac 12 \right)} \\ & = \frac {16}{35} \approx \boxed{0.457} \end{aligned}$