Question Number 128417 by liberty last updated on 07/Jan/21
$$\:\:\:\:\:\:\:\:\:\eta\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{dx}\: \\ $$
Answered by Dwaipayan Shikari last updated on 07/Jan/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{3}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{{n}−\mathrm{1}} {dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{u}\right)^{{n}−\mathrm{1}} {du}\:=\frac{\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\Gamma\left({n}\right)}{\mathrm{3}\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}+{n}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left({n}\right)}{\mathrm{9}\Gamma\left({n}+\frac{\mathrm{4}}{\mathrm{3}}\right)} \\ $$