Question Number 125313 by john_santu last updated on 10/Dec/20
$$\:\:\:\beta\left({x}\right)=\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$
Answered by john_santu last updated on 10/Dec/20
$${let}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} \:=\:\mathrm{1}−{w}^{\mathrm{2}} \\ $$$$\:{x}\:{dx}\:=\:−{w}\:{dw}\: \\ $$$$\beta\left({x}\right)=\:\int\:\frac{\left(\mathrm{1}−{w}^{\mathrm{2}} \right)\left(−{w}\:{dw}\right)}{{w}} \\ $$$$\:=\:\int\:\left({w}^{\mathrm{2}} −\mathrm{1}\right)\:{dw}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{w}^{\mathrm{3}} −{w}\:+\:{c} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{3}}{w}\left({w}^{\mathrm{2}} −\mathrm{3}\right)\:+\:{c} \\ $$$$\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\:\left(\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{3}\right)+\:{c} \\ $$$$\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\:\left(−\mathrm{2}−{x}^{\mathrm{2}} \right)\:+\:{c}\: \\ $$
Answered by liberty last updated on 10/Dec/20
$${we}\:{substitute}\:{x}^{\mathrm{2}} \:=\:{r}\:{then}\:{x}\:=\:{r}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\beta\left({x}\right)=\:\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:=\:\int\:{r}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{r}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}{r}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)\:{dr} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:{r}\left(\mathrm{1}−{r}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dr}\: \\ $$$${let}\:\left(\mathrm{1}−{r}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:{h}\:,\:{r}=\mathrm{1}−{h}^{\mathrm{2}} \:;\:{dr}\:=\:−\mathrm{2}{h}\:{dh} \\ $$$$\beta\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−{h}^{\mathrm{2}} \right){h}^{−\mathrm{1}} \left(−\mathrm{2}{h}\right)\:{dh}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}{h}^{\mathrm{3}} −{h}\:+\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left({h}^{\mathrm{2}} −\mathrm{3}\right)+{c} \\ $$$$=\:\frac{\sqrt{\mathrm{1}−{r}}}{\mathrm{3}}\:\left(−{r}−\mathrm{2}\right)\:+\:{c}\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\left(−{x}^{\mathrm{2}} −\mathrm{2}\right)\:+\:{c} \\ $$
Answered by Ñï= last updated on 10/Dec/20
Commented by Ñï= last updated on 10/Dec/20
$${x}={itan}\theta \\ $$