Question Number 198030 by hardmath last updated on 08/Oct/23
Answered by Mathspace last updated on 09/Oct/23
)),(√(arcsinu))[/ I=(1/(sinc))∫_(√(arcsin(√u))) ^(√(arcsinu)) (√u)du =(1/(sinc))×[(2/3)u^(3/2) ]_(√(arcsin(√u))) ^(√(arcsinu)) =(2/(3sinc)){((√(arcsinu)))^(3/2) −((√(arcsin(√u))))^(3/2) } ⇒lim_(t→1^− ) I(t)=(1/(sin((π/2))))((√((π/2))^(3/2) ))−(√(((π/2))^(3/2) ))} =0 (t→1 ⇒u→1)](https://www.tinkutara.com/question/Q198058.png)
$${let}\:\sqrt{{arcsinx}}={t}\:\Rightarrow{x}={sin}\left({t}^{\mathrm{2}} \right) \\ $$$${and}\:{I}=\int_{\sqrt{{arcsint}}} ^{\sqrt{{arcsin}\left({t}^{\mathrm{2}} \right)}} \:\:\:\frac{{t}}{{sin}\left({t}^{\mathrm{2}} \right)\left({sin}\left({t}^{\mathrm{2}} \right)−\mathrm{1}\right)}\mathrm{2}{tcos}\left({t}^{\mathrm{2}} \right){dt} \\ $$$$=−\mathrm{2}\int_{\sqrt{{arcsint}}} ^{\sqrt{{arcsin}\left({t}^{\mathrm{2}} \right)}} \:\:\frac{{t}^{\mathrm{2}} }{{sin}\left({t}^{\mathrm{2}} \right){cos}\left({t}^{\mathrm{2}} \right)}{cos}\left({t}^{\mathrm{2}} \right){dt} \\ $$$$=\mathrm{2}\int_{\sqrt{{arcsint}}} ^{\sqrt{{arcsint}^{\mathrm{2}} }} \:\:\:\frac{{t}^{\mathrm{2}} }{{sin}\left({t}^{\mathrm{2}} \right)}{dt}\:\:\:\left({t}=\sqrt{{u}}\right) \\ $$$$=\mathrm{2}\int_{\sqrt{{arcsin}\sqrt{{u}}}} ^{\sqrt{{arcsinu}}} \:\:\:\frac{{u}}{{sinu}}×\frac{{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$$=\int_{\sqrt{{arcsin}\sqrt{{u}}}} ^{\sqrt{{arcsinu}}} \:\:\:\frac{\sqrt{{u}}}{{sinu}}{du} \\ $$$$\left.\exists{c}\:\in\right]\sqrt{{arcsin}\sqrt{{u}}},\sqrt{{arcsinu}}\left[/\right. \\ $$$${I}=\frac{\mathrm{1}}{{sinc}}\int_{\sqrt{{arcsin}\sqrt{{u}}}} ^{\sqrt{{arcsinu}}} \:\:\:\sqrt{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{{sinc}}×\left[\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\sqrt{{arcsin}\sqrt{{u}}}} ^{\sqrt{{arcsinu}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}{sinc}}\left\{\left(\sqrt{{arcsinu}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\sqrt{{arcsin}\sqrt{{u}}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\} \\ $$$$ \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{1}^{−} } \:\:{I}\left({t}\right)=\frac{\mathrm{1}}{{sin}\left(\frac{\pi}{\mathrm{2}}\right)}\left(\sqrt{\left.\frac{\pi}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }−\sqrt{\left(\frac{\pi}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right\} \\ $$$$=\mathrm{0}\:\:\:\:\left({t}\rightarrow\mathrm{1}\:\Rightarrow{u}\rightarrow\mathrm{1}\right) \\ $$$$ \\ $$