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Question Number 168231 by naka3546 last updated on 06/Apr/22

$$\int x\sqrt{1-x^6}dx=?$$

Commented by MJS_new last updated on 06/Apr/22

$$\mathrm{first}\mathrm{step}$$$$\int x\sqrt{1-x^6}dx=$$$$[t=x^2\to dx=\frac{dt}{2x}]$$$$=\frac{1}{2}\int \sqrt{1-t^3}dt$$$$\mathrm{this}\mathrm{leads}\mathrm{to}\mathrm{a}\mathrm{hypergeometrical}\mathrm{solution}$$

Commented by naka3546 last updated on 06/Apr/22

$$\mathrm{thank}\mathrm{you},\mathrm{sir}.$$

Commented by MJS_new last updated on 07/Apr/22

$$\frac{1}{2}\int \sqrt{1-t^3}dt=$$$$[u=\arcsin \sqrt{t^3}\to dt=\frac{2\sqrt{1-t^3}}{3\sqrt{t}}du]$$$$=\frac{1}{3}\int {\sin }^{-\frac{1}{3}}u{\cos }^{2}udu=$$$$=\frac{1}{2}{\sin }^{\frac{2}{3}}u{}_2{\mathrm{F}}_1(-\frac{1}{2},\frac{1}{3};\frac{4}{3};{\sin }^{2}u)=$$$$=\frac{1}{2}t{}_2{\mathrm{F}}_1(-\frac{1}{2},\frac{1}{3};\frac{4}{3};t^3)=$$$$=\frac{1}{2}{x}^2{}_2{\mathrm{F}}_1(-\frac{1}{2},\frac{1}{3};\frac{4}{3};x^6)+C$$

Commented by peter frank last updated on 07/Apr/22

$$\mathrm{thanks}$$

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