Question Number 36752 by prof Abdo imad last updated on 05/Jun/18
$${find}\:\:\int\:\:\:\frac{{dx}}{{arcsinx}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18
$${is}\:{it}\:\int\frac{{dx}}{{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right)}\:\:{pls}\:{clarify} \\ $$
Commented by abdo.msup.com last updated on 05/Jun/18
$${find}\:\int\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{arcsin}\left({x}\right)} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18
$${ok}…{sir} \\ $$
Commented by prof Abdo imad last updated on 06/Jun/18
$${changement}\:{x}={sin}\theta\:{give}\: \\ $$$${I}\:=\:\int\:\:\:\:\:\frac{{cos}\theta\:{d}\theta}{\theta\:{cos}\theta}\:=\:\int\:\:\frac{{d}\theta}{\theta}\:={ln}\mid\theta\mid\:+{c} \\ $$$${I}\:=\:{ln}\mid\:{arcsin}\left({x}\right)\mid\:+{c}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18
$$\int\frac{{dx}}{\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right){sin}^{−\mathrm{1}} {x}} \\ $$$${t}={sin}^{−\mathrm{1}} {x} \\ $$$${x}={sint} \\ $$$${dx}={cost}\:{dt} \\ $$$$\int\frac{{cost}\:{dt}}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}\:×{t}} \\ $$$$\int\frac{{dt}}{{t}}={lnt}\:+{c} \\ $$$$={ln}\mid\left({sin}^{−\mathrm{1}} {x}\right)\mid\:+{c} \\ $$