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Question Number 62342 by maxmathsup by imad last updated on 20/Jun/19

let f(ξ) =∫ (x^2 /(√(1−ξx^2 )))dx with 0<ξ<1 1) determine a explicit form of f(ξ) 2) calculate lim_(ξ→1) f(ξ) 3) calculate ∫_0 ^(1/2) (x^2 /(√(1−sin^2 θ x^2 ))) dx with 0<θ<(π/2)

$${let}\:{f}\left(\xi\right)\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }}{dx}\:\:\:{with}\:\:\mathrm{0}<\xi<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left(\xi\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{\xi\rightarrow\mathrm{1}} \:\:\:{f}\left(\xi\right) \\ $$ $$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }}\:{dx}\:{with}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$

Commented byprof Abdo imad last updated on 20/Jun/19

1) f(ξ)=(1/ξ) ∫ ((((√ξ)x)^2 )/(√(1−((√ξ)x)^2 ))) dx =_((√ξ)x=t) (1/ξ) ∫ (t^2 /(√(1−t^2 ))) (dt/(√ξ)) =(1/(ξ(√ξ))) ∫ (t^2 /(√(1−t^2 ))) dt ∫ (t^2 /(√(1−t^2 ))) dt =−∫ ((1−t^2 −1)/(√(1−t^2 )))dt =−∫ (√(1−t^2 ))dt +∫ (dt/(√(1−t^2 ))) +c ∫ (dt/(√(1−t^2 ))) dt =arcsint ∫ (√(1−t^2 ))dt =_(t=sinu) ∫ cos^2 udu =∫ ((1+cos(2u))/2) du =(u/2) +(1/4)sin(2u) =(u/2) +(1/2)sinu cosu =(u/2) +(1/2)sinu(√(1−sin^2 u)) =((arcsint)/2) +(t/2)(√(1−t^2 )) ⇒ f(ξ) =(1/(ξ(√ξ)))(−((arcsint)/2) +(1/2)t(√(1−t^2 )) +arcsint} +c f(ξ)=(1/(ξ(√ξ))){((arcsint)/2) +((t(√(1−t^2 )))/2)} +c

$$\left.\mathrm{1}\right)\:{f}\left(\xi\right)=\frac{\mathrm{1}}{\xi}\:\int\:\frac{\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }{\sqrt{\mathrm{1}−\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }}\:{dx}\: \\ $$ $$=_{\sqrt{\xi}{x}={t}} \:\:\:\:\frac{\mathrm{1}}{\xi}\:\int\:\frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\frac{{dt}}{\sqrt{\xi}}\:=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\:\int\:\:\frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$ $$\int\:\frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:=−\int\:\frac{\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\: \\ $$ $$=−\int\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:+\int\:\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{c} \\ $$ $$\int\:\:\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:={arcsint}\: \\ $$ $$\int\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=_{{t}={sinu}} \:\:\:\int\:{cos}^{\mathrm{2}} {udu} \\ $$ $$=\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\:{du}\:=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right) \\ $$ $$=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sinu}\:{cosu}\:=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sinu}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {u}} \\ $$ $$=\frac{{arcsint}}{\mathrm{2}}\:+\frac{{t}}{\mathrm{2}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$ $${f}\left(\xi\right)\:=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\left(−\frac{{arcsint}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:+{arcsint}\right\}\:+{c} \\ $$ $${f}\left(\xi\right)=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\left\{\frac{{arcsint}}{\mathrm{2}}\:+\frac{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}}\right\}\:+{c}\: \\ $$ $$ \\ $$

Commented byprof Abdo imad last updated on 20/Jun/19

t =x(√ξ) ⇒ f(ξ) =(1/(2ξ(√ξ))){ arcsin(x(√ξ))+x(√ξ)(√(1−ξx^2 ))} +c

$${t}\:={x}\sqrt{\xi}\:\Rightarrow \\ $$ $${f}\left(\xi\right)\:=\frac{\mathrm{1}}{\mathrm{2}\xi\sqrt{\xi}}\left\{\:{arcsin}\left({x}\sqrt{\xi}\right)+{x}\sqrt{\xi}\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }\right\}\:+{c} \\ $$

Commented byprof Abdo imad last updated on 20/Jun/19

2) lim_(ξ→1) f(ξ) =(1/2)( arcsin(x)+x(√(1−x^2 ))) +c

$$\left.\mathrm{2}\right)\:{lim}_{\xi\rightarrow\mathrm{1}} \:\:\:{f}\left(\xi\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{arcsin}\left({x}\right)+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:+{c} \\ $$

Commented byprof Abdo imad last updated on 20/Jun/19

3) ∫_0 ^(1/2) (x^2 /(√(1−sin^2 θ x^2 )))dx (ξ=sin^2 θ) =[(1/(2sin^3 θ)){arcsin(xsinθ) +xsinθ(√(1−sin^2 θ x^2 ))}]_0 ^(1/2) =(1/(2sin^3 θ)){ arcsin(((sinθ)/2))+(1/2)sinθ(√(1−(1/4)sin^2 θ))}

$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }}{dx}\:\:\:\:\:\:\:\:\:\:\left(\xi={sin}^{\mathrm{2}} \theta\right) \\ $$ $$=\left[\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{3}} \theta}\left\{{arcsin}\left({xsin}\theta\right)\:+{xsin}\theta\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }\right\}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{3}} \theta}\left\{\:{arcsin}\left(\frac{{sin}\theta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{sin}\theta\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \theta}\right\} \\ $$

Answered by mr W last updated on 20/Jun/19

f(ξ) =∫ (x^2 /(√(1−ξx^2 )))dx =(1/ξ)∫ ((((√ξ)x)^2 )/(√(1−((√ξ)x)^2 )))d((√ξ)x) =(1/ξ)∫ (t^2 /(√(1−t^2 )))dt =(1/ξ)∫[ (1/(√(1−t^2 )))−(√(1−t^2 ))]dt =.... =(1/(2ξ))[(1/(√ξ)) sin^(−1) ((√ξ)x)−x(√(1−ξx^2 ))]+C

$${f}\left(\xi\right)\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }}{dx} \\ $$ $$=\frac{\mathrm{1}}{\xi}\int\:\:\frac{\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }{\sqrt{\mathrm{1}−\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }}{d}\left(\sqrt{\xi}{x}\right) \\ $$ $$=\frac{\mathrm{1}}{\xi}\int\:\:\frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$ $$=\frac{\mathrm{1}}{\xi}\int\left[\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right]{dt} \\ $$ $$=.... \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}\xi}\left[\frac{\mathrm{1}}{\sqrt{\xi}}\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\xi}{x}\right)−{x}\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }\right]+{C} \\ $$

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