Question Number 168231 by naka3546 last updated on 06/Apr/22
$$\int\:\:{x}\:\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }\:\:{dx}\:\:=\:\:? \\ $$
Commented by MJS_new last updated on 06/Apr/22
![first step ∫x(√(1−x^6 ))dx= [t=x^2 → dx=(dt/(2x))] =(1/2)∫(√(1−t^3 ))dt this leads to a hypergeometrical solution](https://www.tinkutara.com/question/Q168239.png)
$$\mathrm{first}\:\mathrm{step} \\ $$$$\int{x}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }{dt} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{hypergeometrical}\:\mathrm{solution} \\ $$
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
![(1/2)∫(√(1−t^3 ))dt= [u=arcsin (√t^3 ) → dt=((2(√(1−t^3 )))/(3(√t)))du] =(1/3)∫sin^(−(1/3)) u cos^2 u du= =(1/2)sin^(2/3) u _2 F_1 (−(1/2), (1/3); (4/3); sin^2 u) = =(1/2)t _2 F_1 (−(1/2), (1/3); (4/3); t^3 ) = =(1/2)x^2 _2 F_1 (−(1/2), (1/3); (4/3); x^6 ) +C](https://www.tinkutara.com/question/Q168255.png)
$$\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsin}\:\sqrt{{t}^{\mathrm{3}} }\:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }}{\mathrm{3}\sqrt{{t}}}{du}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{3}}} {u}\:\mathrm{cos}^{\mathrm{2}} {u}\:{du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\frac{\mathrm{2}}{\mathrm{3}}} {u}\:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:\mathrm{sin}^{\mathrm{2}} {u}\right)\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{t}\:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:{t}^{\mathrm{3}} \right)\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{4}}{\mathrm{3}};\:{x}^{\mathrm{6}} \right)\:+{C} \\ $$
Commented by peter frank last updated on 07/Apr/22
$$\mathrm{thanks} \\ $$