Question Number 62342 by maxmathsup by imad last updated on 20/Jun/19
$${let}\:{f}\left(\xi\right)\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }}{dx}\:\:\:{with}\:\:\mathrm{0}<\xi<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left(\xi\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{\xi\rightarrow\mathrm{1}} \:\:\:{f}\left(\xi\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }}\:{dx}\:{with}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$
Commented by prof Abdo imad last updated on 20/Jun/19
$$\left.\mathrm{1}\right)\:{f}\left(\xi\right)=\frac{\mathrm{1}}{\xi}\:\int\:\frac{\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }}\:{dx}\: \\ $$$$=_{\sqrt{\xi}{x}={t}} \:\:\:\:\frac{\mathrm{1}}{\xi}\:\int\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\frac{{dt}}{\:\sqrt{\xi}}\:=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\:\int\:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$$$\int\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:=−\int\:\frac{\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\: \\ $$$$=−\int\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:+\int\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{c} \\ $$$$\int\:\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:={arcsint}\: \\ $$$$\int\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=_{{t}={sinu}} \:\:\:\int\:{cos}^{\mathrm{2}} {udu} \\ $$$$=\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\:{du}\:=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right) \\ $$$$=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sinu}\:{cosu}\:=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sinu}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {u}} \\ $$$$=\frac{{arcsint}}{\mathrm{2}}\:+\frac{{t}}{\mathrm{2}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left(\xi\right)\:=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\left(−\frac{{arcsint}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:+{arcsint}\right\}\:+{c} \\ $$$${f}\left(\xi\right)=\frac{\mathrm{1}}{\xi\sqrt{\xi}}\left\{\frac{{arcsint}}{\mathrm{2}}\:+\frac{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}}\right\}\:+{c}\: \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 20/Jun/19
$${t}\:={x}\sqrt{\xi}\:\Rightarrow \\ $$$${f}\left(\xi\right)\:=\frac{\mathrm{1}}{\mathrm{2}\xi\sqrt{\xi}}\left\{\:{arcsin}\left({x}\sqrt{\xi}\right)+{x}\sqrt{\xi}\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }\right\}\:+{c} \\ $$
Commented by prof Abdo imad last updated on 20/Jun/19
$$\left.\mathrm{2}\right)\:{lim}_{\xi\rightarrow\mathrm{1}} \:\:\:{f}\left(\xi\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{arcsin}\left({x}\right)+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:+{c} \\ $$
Commented by prof 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))}](https://www.tinkutara.com/question/Q62382.png)
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }}{dx}\:\:\:\:\:\:\:\:\:\:\left(\xi={sin}^{\mathrm{2}} \theta\right) \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{3}} \theta}\left\{{arcsin}\left({xsin}\theta\right)\:+{xsin}\theta\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta\:{x}^{\mathrm{2}} }\right\}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{3}} \theta}\left\{\:{arcsin}\left(\frac{{sin}\theta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{sin}\theta\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \theta}\right\} \\ $$
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](https://www.tinkutara.com/question/Q62361.png)
$${f}\left(\xi\right)\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\xi}\int\:\:\frac{\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\left(\sqrt{\xi}{x}\right)^{\mathrm{2}} }}{d}\left(\sqrt{\xi}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\xi}\int\:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{\mathrm{1}}{\xi}\int\left[\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right]{dt} \\ $$$$=…. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\xi}\left[\frac{\mathrm{1}}{\:\sqrt{\xi}}\:\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\xi}{x}\right)−{x}\sqrt{\mathrm{1}−\xi{x}^{\mathrm{2}} }\right]+{C} \\ $$