$$ \displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}} \left(1-\left(1-x^3\right)^\sqrt{2}\right)^\sqrt{3}x^2\,\cssId{int-var-mathjax}{\mathrm{d}x} $$
I have been taught Gamma function and Beta function recently. But I cannot evaluate this integral. I am unable to proceed. I need help with the approach.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}