Use the Binomial Expansion If You're Crazy

Noticing the monstrous $1 - x^3$ term inside the power of 10, we try the substitution $u = x^3,$ which leads to $du = 3x^2dx.$ This converts our integral accordingly:

$\int_0^1 x^5 (1 - x^3)^{10}dx \longrightarrow \frac{1}{3} \int_0^1 u(1 - u)^{10}du.$

At this point, we have two options.

Option 1: Continue by making the substitution $v = 1 - u,$ thereby transforming the integral again:

$\frac{1}{3} \int_0^1 u(1 - u)^{10}du \longrightarrow -\frac{1}{3} \int_1^0 (1 - v)v^{10}dv = \frac{1}{3} \int_0^1 v^{10} - v^{11}dv = \boxed{\frac{1}{396}}$

Option 2: Recognize that the integral in terms of $u$ is equivalent to the value of one-third the beta function $B(2, 11) = \frac{1}{3} \times \frac{1! \times 10!}{12!} = \boxed{\frac{1}{396}}.$

sub $u=x^3\to x=u^{1/3}\to dx=\dfrac{u^{-2/3}}{3} du$ boundaries dont change. put this in $\int_0^1 u^{5/3} (1-u)^{10} \dfrac{u^{-2/3}}{3} du=\dfrac{1}{3}\int_0^1 u(1-u)^{10} du\\ =\dfrac{1}{3}B(2,11)=\dfrac{1}{396}$

• substitute 1- $x^{3}$ as $t^{\frac{1}{2}}$

• dx = ${\frac{-1}{6x^{3}}}$ $t^{\frac{-1}{2}}$ dt

• $\int _{ 0 }^{ 1 }{ { x }^{ 5 }{ (1\quad -\quad { x }^{ 3 }) }^{ 10 } } dx$ = $\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ { t }^{ { 9 }/{ 2 } } } (1-{ t }^{ { 1 }/{ 2 } })dt$

Using basic integration and applying limits the above expression evaluates to

$\frac { 1 }{ 6 } (\frac { 2\quad { 1 }^{ { 11 }/{ 2 } } }{ 11 } -\frac { { 1 }^{ 6 } }{ 6 } )$ = $\frac { 1 }{ 396 }$

This is what I did, but not quite as efficient as some of solutions already posted. Notice that $x^3 = \sqrt{x^6}$ . By letting $u = x^6$ , then $du = 6x^5 dx$ and $\frac{1}{6}du = x^5 dx$ and this transforms the integral to

$\frac{1}{6} \int_0^1 (1-\sqrt{u})^{10} du$ . Make another substitution: $v = 1-\sqrt{u}$ , then $dv = \frac{-1}{2\sqrt{u}} du$ and solving for $du$ gives $du = -2\sqrt{u} dv$ . Since $2v-2 = -2\sqrt{u}$ the integral now becomes

$\frac{1}{6} \int_1^0 v^{10}(2v-2)dv = \frac{1}{3} \int_0^1 (v^{10} - v^{11}) dv = \frac{1}{396}$ . Therefore, the answer is 397.

$\displaystyle \int_0^1 x^3 (1-x^3)^{10}\ \ \underbrace{\Big( x^2\,dx\Big)}_\text{HINT!} = \int_0^1 u(1-u)^{10}\left(\frac 1 3 \, du\right) = \frac 1 3 \int_1^0 (1-w) w^{10} \, (-dw).$

$\displaystyle = \frac 1 3 \int_0^1 (w^{10} - w^{11})\,dw = \frac 1 3 \left( \frac 1 {11} - \frac 1 {12} \right) = \frac 1 {132}.$

$\int x^5(1-x^3)^10 = \frac{1}{3}\left(\frac{1}{12}x^{36}-\frac{10}{11}x^{33}+\frac{9}{2}x^{30}-\frac{40}{3}x^{27}+\frac{105}{4}x^{24}-36x^{21}+35x^{18}-24x^{15}+\frac{45}{4}x^12-\frac{10}{3}x^9+\frac{1}{2}x^6\right)+C$

Do not forget to use: $\int_{a}^{b}\,dx = \lim_{x\to b^-}F((x))- \lim_{x\to a^+}F((x))$

So $\frac{1}{396}$ . We it makes 397 $\because$ of $\frac{a}{b}$

$\LARGE \text{ADIOS!!!}$