$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}\bracks{1 - \pars{1 - x^{3}}^{\root{2}}}^{\root{3}}x^{2}\,\dd x}
\\[5mm] \stackrel{x^{3}\ \mapsto\ y}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - \pars{1 - y}^{\root{2}}}^{\root{3}}\dd y
\\[5mm]
\stackrel{1 - y\ \mapsto\ z}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - z^{\root{2}}}^{\root{3}}\dd z
\\[5mm] \,\,\,\stackrel{\pars{1 - z}^{\root{2}}\ \mapsto\ t}{=}\,\,\,&
{\root{2} \over 6}\int_{0}^{1}t^{\root{3}}\pars{1 - t}^{\root{2}/2 - 1}
\,\dd t
\\[5mm] = &\
\bbx{{\root{2} \over 6}\,\mrm{B}\pars{\root{3} + 1,{\root{2} \over 2}}}
\approx 0.1546
\end{align}
$\ds{\mrm{B}}$ is the
Beta Function.
Let $ k\in\mathbb{N} : $
$$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta}+\int_{\frac{\pi}{2}}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$
Then, making the substitution $ \left\lbrace\begin{aligned}\theta &=\pi-x \\ \mathrm{d}\theta &=-\,\mathrm{d}x \end{aligned}\right. $ in the second term, we get that : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=2\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$
Using the substitution $ \left\lbrace\begin{aligned}u&=\sin^{2}{\theta}\\ \mathrm{d}\theta &=\frac{\mathrm{d}x}{2\sqrt{x}\sqrt{1-x}}\end{aligned}\right. $, we get the following : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{1}{x^{\frac{k-1}{2}}\left(1-x\right)^{-\frac{1}{2}}\,\mathrm{d}x}=\beta\left(\frac{k+1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)} $$
Best Answer
The incomplete Beta function is defined $$ B(z;a,b)=\int_0^z u^{a-1}(1-u)^{b-1}du $$ therefore your integral is $$ B(1-\epsilon;2-\frac{\theta}{\mu},n+1) $$ If $\epsilon=0$ you have the normal Beta function composed of Gamma functions, $$ \frac{\Gamma (2-\frac{\theta}{\mu})\Gamma (n+1)}{\Gamma (3 +n-\frac{\theta}{\mu})} $$ To approximate the integral you can expand the incomplete Beta functions in powers of $(1-\epsilon)$ $$ B(1-\epsilon;2-\frac{\theta}{\mu},n+1)=(1-\epsilon)^{2-\frac{\theta}{\mu}}\sum_{k=0}^{\infty}\frac{(\frac{\theta}{\mu}-1)_k}{k!(n+k+1)}(1-\epsilon)^k $$where $(x)_k$ is the Pochhammer symbol.