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Question Number 125313 by john_santu last updated on 10/Dec/20

β(x)=∫ (x^3 /( (√(1−x^2 )))) dx

$$\:\:\:\beta\left({x}\right)=\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$

Answered by john_santu last updated on 10/Dec/20

let (√(1−x^2 )) = w ⇒x^2 = 1−w^2 x dx = −w dw β(x)= ∫ (((1−w^2 )(−w dw))/w) = ∫ (w^2 −1) dw = (1/3)w^3 −w + c = (1/3)w(w^2 −3) + c = ((√(1−x^2 ))/3) (1−x^2 −3)+ c = ((√(1−x^2 ))/3) (−2−x^2 ) + c

$${let}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} \:=\:\mathrm{1}−{w}^{\mathrm{2}} \\ $$$$\:{x}\:{dx}\:=\:−{w}\:{dw}\: \\ $$$$\beta\left({x}\right)=\:\int\:\frac{\left(\mathrm{1}−{w}^{\mathrm{2}} \right)\left(−{w}\:{dw}\right)}{{w}} \\ $$$$\:=\:\int\:\left({w}^{\mathrm{2}} −\mathrm{1}\right)\:{dw}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{w}^{\mathrm{3}} −{w}\:+\:{c} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{3}}{w}\left({w}^{\mathrm{2}} −\mathrm{3}\right)\:+\:{c} \\ $$$$\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\:\left(\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{3}\right)+\:{c} \\ $$$$\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\:\left(−\mathrm{2}−{x}^{\mathrm{2}} \right)\:+\:{c}\: \\ $$

Answered by liberty last updated on 10/Dec/20

we substitute x^2 = r then x = r^(1/2) β(x)= ∫ (x^3 /( (√(1−x^2 )))) dx = ∫ r^(3/2) (1−r)^(−(1/2)) ((1/2)r^(−(1/2)) ) dr = (1/2)∫ r(1−r)^(−(1/2)) dr let (1−r)^(1/2) = h , r=1−h^2 ; dr = −2h dh β(x) = (1/2)∫(1−h^2 )h^(−1) (−2h) dh = (1/3)h^3 −h + c = (1/3)h(h^2 −3)+c = ((√(1−r))/3) (−r−2) + c = ((√(1−x^2 ))/3)(−x^2 −2) + c

$${we}\:{substitute}\:{x}^{\mathrm{2}} \:=\:{r}\:{then}\:{x}\:=\:{r}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\beta\left({x}\right)=\:\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:=\:\int\:{r}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{r}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}{r}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)\:{dr} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:{r}\left(\mathrm{1}−{r}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dr}\: \\ $$$${let}\:\left(\mathrm{1}−{r}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:{h}\:,\:{r}=\mathrm{1}−{h}^{\mathrm{2}} \:;\:{dr}\:=\:−\mathrm{2}{h}\:{dh} \\ $$$$\beta\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−{h}^{\mathrm{2}} \right){h}^{−\mathrm{1}} \left(−\mathrm{2}{h}\right)\:{dh}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}{h}^{\mathrm{3}} −{h}\:+\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left({h}^{\mathrm{2}} −\mathrm{3}\right)+{c} \\ $$$$=\:\frac{\sqrt{\mathrm{1}−{r}}}{\mathrm{3}}\:\left(−{r}−\mathrm{2}\right)\:+\:{c}\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}\left(−{x}^{\mathrm{2}} −\mathrm{2}\right)\:+\:{c} \\ $$

Answered by Ñï= last updated on 10/Dec/20

Commented by Ñï= last updated on 10/Dec/20

x=itanθ

$${x}={itan}\theta \\ $$

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